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Usually, an infinity in a solution to an equation in physics is telling us that the equation breaks down, and must be replaced with something else. In the examples you give about fluid mechanics and Euler disks, we certainly don't expect real fluids and real disks to blow up in finite time. If such behavior is observed mathematically, it is a sign that the equations used to derive those solutions are not good models of the physical situation in that regime.

In your situation, you have drawn a curve which could be a trajectory that a particle follows. Let's forget about special relativity for a moment, which just complicates things here (I'll touch on it at the end of the answer). In order for it to be a possible solution to the equations of classical physics, that curve $x(t)$ should satisfy Newton's second law \begin{equation} F = m \frac{d^2x}{dt^2} \end{equation} Let's evaluate the right hand side for your proposed motion (by the way, as a small correction, the argument of the log should always be dimensionless, so we should add a constant $t_0$ to make this work, and to make the dimensions work out, I will also multiply your expression by $x(t)$ by a constant $x_0$ with units of length) \begin{equation} m\frac{d^2}{dt^2} \left[x_0 \frac{t}{2 t_0} \log \left(\frac{t^2}{t_0^2}\right)\right] = \frac{m x_0}{t_0} \frac{1}{t} \end{equation} Now as $t\rightarrow 0$, we see that your particle has an infinite acceleration. This requires that the force applied to this particle is also infinite. (As an aside, this also makes this case less mathematically interesting than fluid mechanics -- of course if we put in an infinite force, than the solution of the equation will blow up. The case of fluid mechanics is interesting because from apparently innocuous initial conditions, the solution's natural evolution may cause it to blow up, without us having to putting that blow up in by hand.)

Realistically, we expect that any physical force acting on your particle will be bounded by some finite amount. So this is not a plausible motion. However, you can have motions that are arbitrarily close to the one you proposed, with a finite force. For example, suppose your motion was actually \begin{equation} x(t) = \begin{cases} x_0 \frac{t}{2 t_0} \log\left(\frac{t^2}{t_0^2}\right), & |t| \geq \epsilon \\ p_\epsilon(t), & |t| < \epsilon \\ \end{cases} \end{equation} for some $\epsilon>0$, and where $p_\epsilon(t)$ is some polynomial which is chosen so that $x(t)$ is continuous and at least twice differentiable at $t = \pm \epsilon$. Then, your motion will have a finite acceleration throughout the trajectory. By taking $\epsilon$ arbitrarily close to $0$, we can make your curve arbitrarily close to the one you wanted, with a finite force. This explains why when you look at a plot of your $x(t)$, the motion is apparently fine. It is hard to see from the curve the one point where the solution is blowing up, and curves that are arbitrarily close to the one you drew are ok (in classical physics).

If you want to generalize this situation to special relativity, the equations of motion of classical physics get modified, in such a way that is not possible to find a solution where the velocity accelerates from a value below $c$, to $c$ or a value above $c$, without expending an infinite amount of energy. Again, this infinity is a sign that something is wrong, physically -- in the case of special relativity, the problem is that massive particles must travel slower than $c$.

Usually, an infinity in a solution to an equation in physics is either telling us that the equation breaks down, and must be replaced with something else, or that you made an unphysical assumption in deriving the solution. In the examples you give about fluid mechanics and Euler disks, we certainly don't expect real fluids and real disks to blow up in finite time. If such behavior is observed mathematically, it is a sign that the equations used to derive those solutions are not good models of the physical situation in that regime.

In your situation, you have drawn a curve which could be a trajectory that a particle follows. Let's forget about special relativity for a moment, which just complicates things here (I'll touch on it at the end of the answer). In order for it to be a possible solution to the equations of classical physics, that curve $x(t)$ should satisfy Newton's second law \begin{equation} F = m \frac{d^2x}{dt^2} \end{equation} Let's evaluate the right hand side for your proposed motion (by the way, as a small correction, the argument of the log should always be dimensionless, so we should add a constant $t_0$ to make this work, and to make the dimensions work out, I will also multiply your expression by $x(t)$ by a constant $x_0$ with units of length) \begin{equation} m\frac{d^2}{dt^2} \left[x_0 \frac{t}{2 t_0} \log \left(\frac{t^2}{t_0^2}\right)\right] = \frac{m x_0}{t_0} \frac{1}{t} \end{equation} Now as $t\rightarrow 0$, we see that your particle has an infinite acceleration. This requires that the force applied to this particle is also infinite. (As an aside, this also makes this case less mathematically interesting than fluid mechanics -- of course if we put in an infinite force, than the solution of the equation will blow up. The case of fluid mechanics is interesting because from apparently innocuous initial conditions, the solution's natural evolution may cause it to blow up, without us having to putting that blow up in by hand.)

Realistically, we expect that any physical force acting on your particle will be bounded by some finite amount. So this is not a plausible motion. However, you can have motions that are arbitrarily close to the one you proposed, with a finite force. For example, suppose your motion was actually \begin{equation} x(t) = \begin{cases} x_0 \frac{t}{2 t_0} \log\left(\frac{t^2}{t_0^2}\right), & |t| \geq \epsilon \\ p_\epsilon(t), & |t| < \epsilon \\ \end{cases} \end{equation} for some $\epsilon>0$, and where $p_\epsilon(t)$ is some polynomial which is chosen so that $x(t)$ is continuous and at least twice differentiable at $t = \pm \epsilon$. Then, your motion will have a finite acceleration throughout the trajectory. By taking $\epsilon$ arbitrarily close to $0$, we can make your curve arbitrarily close to the one you wanted, with a finite force. This explains why when you look at a plot of your $x(t)$, the motion is apparently fine. It is hard to see from the curve the one point where the solution is blowing up, and curves that are arbitrarily close to the one you drew are ok (in classical physics).

If you want to generalize this situation to special relativity, the equations of motion of classical physics get modified, in such a way that is not possible to find a solution where the velocity accelerates from a value below $c$, to $c$ or a value above $c$, without expending an infinite amount of energy accelerating the particle. Again, this infinity is a sign that something is wrong, physically -- in the case of special relativity, the problem is that massive particles must travel slower than $c$.

Usually, an infinity in a solution to an equation in physics is telling us that the equation breaks down, and must be replaced with something else. In the examples you give about fluid mechanics and Euler disks, we certainly don't expect real fluids and real disks to blow up in finite time. If such behavior is observed mathematically, it is a sign that the equations used to derive those solutions are not good models of the physical situation in that regime.

In your situation, you have drawn a curve which could be a trajectory that a particle follows. Let's forget about special relativity for a moment, which just complicates things here (I'll touch on it at the end of the answer). In order for it to be a possible solution to the equations of classical physics, that curve $x(t)$ should satisfy Newton's second law \begin{equation} F = m \frac{d^2x}{dt^2} \end{equation} Let's evaluate the right hand side for your proposed motion (by the way, as a small correction, the argument of the log should always be dimensionless, so we should add a constant $t_0$ to make this work, and to make the dimensions work out, I will also multiply your expression by $x(t)$ by a constant $x_0$ with units of length) \begin{equation} m\frac{d^2}{dt^2} \left[x_0 \frac{t}{2 t_0} \log \left(\frac{t^2}{t_0^2}\right)\right] = \frac{m x_0}{t_0} \frac{1}{t} \end{equation} Now as $t\rightarrow 0$, we see that your particle has an infinite acceleration. This requires that the force applied to this particle is also infinite.

Realistically, we expect that any physical force acting on your particle will be bounded by some finite amount. So this is not a plausible motion. However, you can have motions that are arbitrarily close to the one you proposed, with a finite force. For example, suppose your motion was actually \begin{equation} x(t) = \begin{cases} x_0 \frac{t}{2 t_0} \log\left(\frac{t^2}{t_0^2}\right), & |t| \geq \epsilon \\ p_\epsilon(t), & |t| < \epsilon \\ \end{cases} \end{equation} for some $\epsilon>0$, and where $p_\epsilon(t)$ is some polynomial which is chosen so that $x(t)$ is continuous and at least twice differentiable at $t = \pm \epsilon$. Then, your motion will have a finite acceleration throughout the trajectory. By taking $\epsilon$ arbitrarily close to $0$, we can make your curve arbitrarily close to the one you wanted, with a finite force.

If you want to generalize this situation to special relativity, the equations of motion of classical physics get modified, in such a way that is not possible to find a solution where the velocity accelerates from a value below $c$, to $c$ or a value above $c$, without expending an infinite amount of energy. Again, this infinity is a sign that something is wrong, physically -- in the case of special relativity, the problem is that massive particles must travel slower than $c$.

Usually, an infinity in a solution to an equation in physics is telling us that the equation breaks down, and must be replaced with something else. In the examples you give about fluid mechanics and Euler disks, we certainly don't expect real fluids and real disks to blow up in finite time. If such behavior is observed mathematically, it is a sign that the equations used to derive those solutions are not good models of the physical situation in that regime.

In your situation, you have drawn a curve which could be a trajectory that a particle follows. Let's forget about special relativity for a moment, which just complicates things here (I'll touch on it at the end of the answer). In order for it to be a possible solution to the equations of classical physics, that curve $x(t)$ should satisfy Newton's second law \begin{equation} F = m \frac{d^2x}{dt^2} \end{equation} Let's evaluate the right hand side for your proposed motion (by the way, as a small correction, the argument of the log should always be dimensionless, so we should add a constant $t_0$ to make this work, and to make the dimensions work out, I will also multiply your expression by $x(t)$ by a constant $x_0$ with units of length) \begin{equation} m\frac{d^2}{dt^2} \left[x_0 \frac{t}{2 t_0} \log \left(\frac{t^2}{t_0^2}\right)\right] = \frac{m x_0}{t_0} \frac{1}{t} \end{equation} Now as $t\rightarrow 0$, we see that your particle has an infinite acceleration. This requires that the force applied to this particle is also infinite. (As an aside, this also makes this case less mathematically interesting than fluid mechanics -- of course if we put in an infinite force, than the solution of the equation will blow up. The case of fluid mechanics is interesting because from apparently innocuous initial conditions, the solution's natural evolution may cause it to blow up, without us having to putting that blow up in by hand.)

Realistically, we expect that any physical force acting on your particle will be bounded by some finite amount. So this is not a plausible motion. However, you can have motions that are arbitrarily close to the one you proposed, with a finite force. For example, suppose your motion was actually \begin{equation} x(t) = \begin{cases} x_0 \frac{t}{2 t_0} \log\left(\frac{t^2}{t_0^2}\right), & |t| \geq \epsilon \\ p_\epsilon(t), & |t| < \epsilon \\ \end{cases} \end{equation} for some $\epsilon>0$, and where $p_\epsilon(t)$ is some polynomial which is chosen so that $x(t)$ is continuous and at least twice differentiable at $t = \pm \epsilon$. Then, your motion will have a finite acceleration throughout the trajectory. By taking $\epsilon$ arbitrarily close to $0$, we can make your curve arbitrarily close to the one you wanted, with a finite force. This explains why when you look at a plot of your $x(t)$, the motion is apparently fine. It is hard to see from the curve the one point where the solution is blowing up, and curves that are arbitrarily close to the one you drew are ok (in classical physics).

If you want to generalize this situation to special relativity, the equations of motion of classical physics get modified, in such a way that is not possible to find a solution where the velocity accelerates from a value below $c$, to $c$ or a value above $c$, without expending an infinite amount of energy. Again, this infinity is a sign that something is wrong, physically -- in the case of special relativity, the problem is that massive particles must travel slower than $c$.

Usually, an infinity in a solution to an equation in physics is telling us that the equation breaks down, and must be replaced with something else. In the examples you give about fluid mechanics and Euler disks, we certainly don't expect real fluids and real disks to blow up in finite time. If such behavior is observed mathematically, it is a sign that the equations used to derive those solutions are not good models of the physical situation in that regime.

In your situation, you have drawn a curve which could be a trajectory that a particle follows. Let's forget about special relativity, which just complicates things here. In order for it to be a possible solution to the equations of classical physics, that curve $x(t)$ should satisfy Newton's second law \begin{equation} F = m \frac{d^2x}{dt^2} \end{equation} Let's evaluate the right hand side for your proposed motion (by the way, as a small correction, the argument of the log should always be dimensionless, so we should add a constant $t_0$ to make this work, and to make the dimensions work out, I will also multiply your expression by $x(t)$ by a constant $x_0$ with units of length) \begin{equation} m\frac{d^2}{dt^2} \left[x_0 \frac{t}{2 t_0} \log \left(\frac{t^2}{t_0^2}\right)\right] = \frac{m x_0}{t_0} \frac{1}{t} \end{equation} Now as $t\rightarrow 0$, we see that your particle has an infinite acceleration. This requires that the force applied to this particle is also infinite.

Realistically, we expect that any physical force acting on your particle will be bounded by some finite amount. So this is not a plausible motion. However, you can have motions that are arbitrarily close to the one you proposed, with a finite force. For example, suppose your motion was actually \begin{equation} x(t) = \begin{cases} x_0 \frac{t}{2 t_0} \log\left(\frac{t^2}{t_0^2}\right), & |t| \geq (1+\epsilon) t_0 \\ p_\epsilon(t), & |t| < (1+\epsilon) t_0 \\ \end{cases} \end{equation} for some $\epsilon>0$, and where $p_\epsilon(t)$ is some polynomial which is chosen so that $x(t)$ is continuous and at least twice differentiable at $\pm (1+\epsilon) t_0$. Then, your motion will have a finite acceleration throughout the trajectory. By taking $\epsilon$ arbitrarily close to $0$, we can make your curve arbitrarily close to the one you wanted, with a finite force.

If you want to generalize this situation to special relativity, the equations of motion of classical physics get modified, in such a way that is not possible to find a solution where the velocity accelerates from a value below $c$, to $c$ or a value above $c$, without expending an infinite amount of energy. Again, this infinity is a sign that something is wrong, physically -- in the case of special relativity, the problem is that massive particles must travel slower than $c$.

Usually, an infinity in a solution to an equation in physics is telling us that the equation breaks down, and must be replaced with something else. In the examples you give about fluid mechanics and Euler disks, we certainly don't expect real fluids and real disks to blow up in finite time. If such behavior is observed mathematically, it is a sign that the equations used to derive those solutions are not good models of the physical situation in that regime.

In your situation, you have drawn a curve which could be a trajectory that a particle follows. Let's forget about special relativity for a moment, which just complicates things here (I'll touch on it at the end of the answer). In order for it to be a possible solution to the equations of classical physics, that curve $x(t)$ should satisfy Newton's second law \begin{equation} F = m \frac{d^2x}{dt^2} \end{equation} Let's evaluate the right hand side for your proposed motion (by the way, as a small correction, the argument of the log should always be dimensionless, so we should add a constant $t_0$ to make this work, and to make the dimensions work out, I will also multiply your expression by $x(t)$ by a constant $x_0$ with units of length) \begin{equation} m\frac{d^2}{dt^2} \left[x_0 \frac{t}{2 t_0} \log \left(\frac{t^2}{t_0^2}\right)\right] = \frac{m x_0}{t_0} \frac{1}{t} \end{equation} Now as $t\rightarrow 0$, we see that your particle has an infinite acceleration. This requires that the force applied to this particle is also infinite.

Realistically, we expect that any physical force acting on your particle will be bounded by some finite amount. So this is not a plausible motion. However, you can have motions that are arbitrarily close to the one you proposed, with a finite force. For example, suppose your motion was actually \begin{equation} x(t) = \begin{cases} x_0 \frac{t}{2 t_0} \log\left(\frac{t^2}{t_0^2}\right), & |t| \geq \epsilon \\ p_\epsilon(t), & |t| < \epsilon \\ \end{cases} \end{equation} for some $\epsilon>0$, and where $p_\epsilon(t)$ is some polynomial which is chosen so that $x(t)$ is continuous and at least twice differentiable at $t = \pm \epsilon$. Then, your motion will have a finite acceleration throughout the trajectory. By taking $\epsilon$ arbitrarily close to $0$, we can make your curve arbitrarily close to the one you wanted, with a finite force.

If you want to generalize this situation to special relativity, the equations of motion of classical physics get modified, in such a way that is not possible to find a solution where the velocity accelerates from a value below $c$, to $c$ or a value above $c$, without expending an infinite amount of energy. Again, this infinity is a sign that something is wrong, physically -- in the case of special relativity, the problem is that massive particles must travel slower than $c$.