The Stack Overflow podcast is back! Listen to an interview with our new CEO.
6

When you write: $$ E = mc^2 \tag{1} $$ the $m$ in the equation is the rest mass, which is an invariant. The speed of light is also an invariant, so the time derivative of the right hand side is zero. The problem is that you are differentiating an equation when both sides are constant. Mathematically you can do this, but physically it isn't very ...


5

Well, $E=mc^2$ doesn't mean that a Joule is equal to a kilogram times (the speed of light) squared. Actually, if a Joule is defined to be $$\operatorname{Joule} =\operatorname{Newton}\cdot \operatorname{meter} = (\operatorname{kg}\cdot \frac{\operatorname{meter}}{\operatorname{second}^2})\cdot \operatorname{meter}=\operatorname{kg}\cdot \left( \frac{\...


4

The assumptions on the SI units are not correct. The kilogram isn't 'defined by an arbitrary weight', it's defined using Planck's constant since 2018: The kilogram is defined by setting the Planck constant h exactly to 6.62607015×10−34 J⋅s (J = kg⋅m2⋅s−2), given the definitions of the metre and the second Hence, the kilogram is now essentially defined ...


3

Perhaps your question is more suited to the psychology stack exchange, but on the assumption that relatively few psychologists will have thought much about SR, perhaps it is worth making the exception to discuss it here. My personal experience was that, having been naturally gifted at physics at school, and despite having a PhD in quantum theory, I found ...


2

Relativists no longer use the relativistic mass convention: http://physics.stackexchange.com/a/133395/4552 But yes, if you're using that convention, then the factor in both cases is $\gamma$. And is time dilation with a specific speed-mass equal to the time dilation with a stationary mass (like a planet). If so, can we say that it is mass alone, and not ...


2

There are two objects in relativity that contain the same information, but are of different mathematical nature - vectors and covectors. Covector is dual vector to the original vector. That is, it is member of different vector space, but (in space endowed with metric) there exists natural one-to-one correspondence between vectors and covectors. Now, since ...


2

GR does not have to be described in a formalism of curved spacetime. You can have, e.g., Deser's spin-2 field on flat spacetime. This theory is inconsistent until you add corrections, which end up making it equivalent to GR. The equivalence principle is what prevents us from deciding whether a theory like Deser's is what "really" happens, as opposed to the ...


2

“Scalar” and “Lorentz invariant” are synonyms in the context of Special and General Relativity. However, it is possible to have constant tensors whose components don’t actually change when transformed, such as the Minkowski metric tensor $\eta_{\mu\nu}$ in Special Relativity. We don’t call these “invariant”. Some people call these tensors “isotropic”; ...


2

I prefer to think of proper time as the 'distance' between the two events in spacetime. Consider some sort of a world-line in a four dimensional spacetime $x^\mu = \left(ct, \mathbf{r}\right)^\mu$ Where $c$ is the speed of light, $t$ is time and $\mathbf{r}$ is position. Lets define some point (event) on this curve as the 'start': $x_0^\mu =\left(ct_0, \...


2

In the Galilean transformation, the velocity $V$ in $r' = r - Vt$ is constant. A derivative of a constant is $0$. In detail: $$ \begin{equation} \frac{d^2}{dt^2} (r-Vt) = \frac{d^2r}{dt^2} - \frac{d} {dt}\frac{d(Vt)}{dt} \end{equation} \\ = \frac{d^2r}{dt^2} - \frac{dV}{dt} \\ = \frac{d^2r}{dt^2} - 0 $$ If $r$ also were a constant, the equation would be ...


2

Short answer: there are no additional factors in this formula because unit of measure for energy is consistent with units of measure for mass, time and distance. Suppose you decided to create your own system of units and let's forget about relativity theory so far. You can choose arbitrary unit for mass. You can choose arbitrary unit for length. You can ...


2

This is completely wrong. The international system of units SI Defines : Time: $1$ second equals: The duration of 9192631770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium-133 atom. Length: $1$ meter equals: The distance travelled by light in vacuum in $(1/299792458) \...


2

Position and center-of-mass coordinates are not quantized/discrete in conventional QM (excluding lattice formulations), cf. e.g. this Phys.SE post. In contrast, for the reason why angular momentum is quantized, see e.g. my Phys.SE answer here.


1

V is the relative speed of the two frames of reference, and must be a constant if its derivative with respect to time is zero. r is the position of the accelerating body, and must be varying if the body is accelerating.


1

You basicaly forgot time dilatation, length contraction and relativity of simultaneity. Since the diagram is drown in the P2 system, it takes 2 seconds for both light signals to reach P2 in this system. When you transform to P1, the speed of P2 is indeed 0,9c, but now the length is contracted by lorentz factor $\gamma$, so the distance in P1 frame is $1.8c*...


1

You're making your life unnecessarily difficult by using three-acceleration and its unique transformation formula. It's easier to use four-vectors for everything, since they transform in a consistent way. But then, you're making your life unnecessarily difficult by solving the same problem in two different frames to begin with – one is enough, since they're ...


1

Every Lorentz invariant is a Lorentz scalar. That is just true by definition. The trick here is to specify what an object is a scalar, vector, tensor, spinor, etc with respect to. For instance, the electric charge density is a scalar with respect to spatial rotations, but not with respect to boosts. Likewise, the four-momentum of a neutral particle can ...


1

Consider two timelike related events. Now draw a worldline for an observer that visits [was present at] both events. The proper time for that worldline is the time elapsed on that observer's wristwatch. In the frame of that observer, those events are in the same position (here at the origin). For another worldline visiting those two events, one gets a ...


1

In general relativity a reference frame need not be inertial. The universe does have a preferred reference frame. It is the one in which the CMB radiation is isotropic to first approximation. This is also the frame in which velocities of galaxies relative to the frame are zero on average. In cosmology it is called the "co-moving frame" and observers at ...


1

This is an arbitrary convention that was fixed historically around the time that Einstein published the general theory of relativity. It's similar to the right-hand rule for defining torque, or the convention that the charge of the electron is negative. Although it's arbitrary, it's fixed, so different authors do not use different conventions as a matter of ...


1

Yes. Both are observed from a frame moving with respect to the frame of the clock and the mass in question. Consider one clock and one mass that stay together, and several observers all moving at different speeds. The observed slowing of time and the increase in mass both depend on the relative speed with which the observer is moving. Observers moving at ...


1

Pulsar's answer was helpful but I want to lay out the misconception more clearly. The setup is to consider the inside of the relativistic rocket with clocks placed at different locations. I jumped to Rindler coordinates as the well-known coordinates to describe events within the rocket. From there, I correctly derive the time difference between clocks at ...


1

This is not a sufficient proof of the spin-statistics theorem because it has little to do with what the spin-statistics theorem says. For one thing, this theorem is a statement about operators, while the property discussed in the OP is purely about classical fields. Let $a$ be an operator that transforms according to a representation $r$ of the Lorentz ...


1

This perturbation operator is a scalar (in the sense that it doesn't change upon rotations). This is the meaning of the fact that it commutes with the angular momentum operators (or that it is spherically symmetrical, as you say). Since it is a scalar, it cannot connect two states with different angular momentum values or different projections of the ...


Only top voted, non community-wiki answers of a minimum length are eligible