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0

For instance, take a rigid pole of several AU in length. [...] The person on the opposite end should receive the pushes and pulls instantaneously as no particle is making the full journey. As other answers have pointed out, you can't have perfect rigidity, and the signal would propagate at the speed of sound in the material. If tap a steel rod with a ...


0

The paradox is: As each twin may consider herself in a distinct inertial reference frame, then each should see the other moving at a relativistic speed, and each should see the other appearing younger than herself, a patently impossible outcome. There is more than one way to resolve this seeming paradox. But regardless of which logical method is used to ...


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Draw a spacetime diagram. Really, there is no better way to solve relativity problems. In the above, the nail has worldlines $1$ (back) and $2$ (front), while the hole's worldlines are $3$ (front) and $4$ (back). Let's agree that the origin $\mathcal{O}$ of the coordinates is the event of the front of the nail just entering the hole, i.e. the intersection ...


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Imagine a slightly different scenario: two pilots, Alice, and Bob, are in their spaceships. They move towards a tunnel of length $L$ at a velocity $v$, and remain a distance $l'$ apart. Alice is closest to the tunnel and thus enters first, approaching a wall at the end of the length of the tunnel. Just as Bob enters he decelerates, coming quickly to a halt ...


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I believe this question is not worded correctly only because the submitter is not that well versed in scientific terminology. To put it briefly - neither sound waves nor light waves have mass, that is because they are representations of moving particles, they themselves are not entities and do not exist beyond the scope of human-facilitating terminology. As ...


-2

This is a variation of the pole and the barn paradox, and is also known as the bug and the rivet paradox, see Rod Nave's hyperphysics: The final line of this article is "the paradox is not resolved". Some people will say the paradox is resolved via consideration of simultaneity, but I don't think it is. Another variation of the theme is when you and I are ...


-2

The bug lives! Since BOTH the hole and the nail will change length in the moving coordinate system and their proportions Ln/Lh will be constant, if you pick a third coordinate system that is moving at half the speed of the nail. In this system both the hole and the nail are moving at the same speed in opposite directions.


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The bug will die: What does cause the nail deceleration? it's the finger at the base of the nail hitting the outside of the hole. In the best case scenario for the bug the tip of nail stops as soon as it gets the information (the shock wave from the abrupt deceleration). This will travel across the nail no faster than the speed of light. So let's assume the ...


1

When considering relativistic speeds, the notion of "particle & anti-particle" somewhat blurs. The correct treatment of a relativistic free electron for example is given by the Dirac Equation, which relates Dirac Spinors. A spinor is something like a 4-vector, describing the wave function of our electron. In it's rest frame, two of the spinor's ...


0

If you define your gamma matrices as the block matrices: $$ \gamma^0:=\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}; \quad \gamma^i:=\begin{bmatrix} 0 & \sigma^i \\ -\sigma^i & 0 \end{bmatrix} $$ Then you can also define the block matrices ...


0

You don't define $\overrightarrow{\sigma}$. It looks like neither do Itzykson/Zuber, or, rather, it is defined there as a vector with Pauli matrices as components (if I am not mistaken). However, those matrices are 2x2, whereas all other matrices in your equation are 4x4. Therefore, such definition seems incompatible with your equation. I believe you should ...


3

You don't need to use relativity to see what will happen in to a current in a gravitational field. We assume a wire of constant cross section with length in the vertical direction , and a constant current flowing through it. This must be done by applying an electric field $\mathbf{E}$ along the length of the wire, producing the current density (according to ...


1

Current is a measure of charge flowing in some unit of time. The gravitational field will dilate time. The current will be reduced in the observer's point of view. $I = V/R$. Since resistance is constant, this means that voltage must have decreased in the observers point of view.


11

Each particle only annihilates its exact antiparticle. Electrons annihilate positrons. A blue up quark annihilates an anti-blue anti-up quark. A muon annihilates an anti-muon. The thing about anti-matter is that it postulates an exact opposite of every particular particle type (except for things like photons that are their own antiparticles). It's about ...


32

A sophisticated, yet easy way to see that this the answer must be "No." is to recall that velocity is relative — that there is no absolute notion of velocity. You said the matter was moving and the antimatter still, but that point of view (AKA frame of reference) is not privileged in any way. An observer at rest with respect to the matter has just as much ...


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The particle-antiparticle annihilation is on a per-particle basis. One electron annihilates on positron. One up quark annihilates one anti-up quark. One down quark annihilates one anti-down quark. Moving at relativistic speeds doesn't change the number of particles. For that matter, you could annihilate an electron with an anti-muon, since an electron and a ...


2

$E=\pm(p_{x}^2+m^2)^{1/2}$ That is true, though. The relativistic energy-momentum relation, $E^2 = m^2 + p^2$. So you can use it. In fact, if you didn't already know about rapidity, this would be a way to discover it: notice that $$E^2 - p^2 = m^2 = E'^2 - p'^2$$ which is reminiscent of the trig identity $$\cosh^2\vartheta - \sinh^2\vartheta = 1 = ...


0

What's wrong with $E = \pm (p_x^2 + m^2)^{1/2}$? As I'm not sure if this qualifies as a homework-related question, I'll just give you a hint: try to write energy and momentum in terms of mass, $\beta$ and $\gamma$. It's easier to solve that way.


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... Another way to look at it is to use geometry which includes the vertical time axis, the horizontal space axis, and the stacking of both motion vectors( c, v ) and length scalars. Rotation determines the direction of travel in space-time, thus determines the velocity across space and the rate of the ticking of time. Here, twin spaceship A is a rest in ...


0

Special relativity itself makes it clear that absolutes cannot be detected, such as being at absolute rest in space, or the inability to detect absolute motion. This therefore prevents one from having an absolute understanding of special relativity, since that which special relativity reveals does not extend to the point of absolutes. Thus the absolute ...


0

Take a look at this pendulum periodicity calculator: http://hyperphysics.phy-astr.gsu.edu/hbase/pend.html. You can compare the period of each swing of the pendulum under different conditions of gravitational acceleration (little g in your formula). Notice that as gravity grows stronger (as gravitational acceleration becomes greater), the period of the ...


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No, the relationship between the period of a pendulum and $g$ is simple Newtonian mechanics and unrelated to special or general relativity. This is discussed in the answers to Time period related to acceleration due to gravity (though I hesitate to link this as that question was not well received). Time dilation was actually known before Einstein formulated ...


1

You probably already understand that merely rotating a coordinate system does not change the physical system you're modeling. Given one set of axes, any rotation of those axes just represents some freedom of choice to do math how you see fit. So with that in mind, if you were pointing northeast (and chose a pair of coordinate axes to point northeast and ...


1

Bernhard Schutz discusses this reasonably well in his book A First Course in General Relativity. Consider sending a light beam horizontally along an $x$ axis and then receiving it back again. A space time plot of this would look like Here's an example of the rotation you are describing And this is how everything becomes distorted when you create such ...


4

Yes, there's a very famous example: muons produced in the upper atmosphere can be detected on the surface of the Earth. Moving at nearly the speed of light, it takes them over 300 microseconds to get down to the Earth's surface, but the average muon decays after 2.2 microseconds. If it were not for time dilation, only a few in every $10^{60}$ muons (so, ...


2

For any isolated object, i.e. one that is having an external force applied to it, the frame that object lives in is locally inertial. By locally I mean that if you consider a small enough region of spacetime around the object it will be impossible to tell the frame isn't inertial. How small this volume is depends on how curved spacetime is. This is an ...


1

The answers are yes and no. Special relativity does make ellipses precess, but it only accounts for 7" out of 43" per century of Mercury's anomalous precession. I wonder if Einstein and/or Sommerfeld knew that. To first order, incorporating special relativity results in a small inverse cube correction to the gravitational force, which is well known to ...


1

Just to complement John Rennie's answer, one can always perform a Lorentz transformation to a coordinate system such as the particle is at rest for a given time. It's called instantaneous rest frame (IRF). This frame changes point to point, unless the particle's velocity is constant. In such a frame, we have $ ds^2 = -c^2d\tau^2, $ where $\tau$ is the ...


1

There is, in a sense, a way to 'guide' oneself to the equations of motion based on the symmetries. The form of mechanics most suitable for this purpose is Hamilton's principle - the system takes a path for which the action has a stationary value for variations with fixed endpoints: $$\delta S=0$$ $S$ is generally expressed as (under some parametrization of ...


0

It depends on how you define mass. I like to think of it as mass is just rest mass. I mean the mass you weight on a scale when nothing is moving. On different media light moves slower not because it gains mass but because of its interaction with the atoms in the media. The photons get absorbed and reemited in such a manner that when you sum the waves for ...


1

First, we will look at the energy of a free relativistic particle of (rest) mass $m$ moving with velocity $v$: $$E = \frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$$ where $E=mc^2$ when $v=0$. We now consider a few cases: $m\ne0$: In this case, $E\rightarrow\infty$ as $v\rightarrow c$. Therefore, a massive particle that at any point of time is moving at less than ...


1

The mistake is that in the equation under "One checks that", the term $-c^2$ shouldn't be there. Instead, one should insert minus the squared value of the $t$-derivative of $c$, and the derivative of a constant is zero. This is assuming that $t$ is really a different name for $s$, an arbitrary parameter along the curve, and the letter $c$ means that the ...


1

Every particle needs to have energy to be a particle (if it had none it wouldn't even exist). Since energy is equivalent to mass and therefore gravitates I would say YES, all particles that have a speed less than the speed of light must also have mass. Because the speed of the particle is less than the speed of light an observer could travel with the same ...


2

I recently read a paper on the possible kinematics: http://scitation.aip.org/content/aip/journal/jmp/9/10/10.1063/1.1664490 It states that under the 3 assumptions they made, there were more then 10 possible Lie-Algebras (while they discarded one by heuristic arguments)


2

The electric and magnetic fields transform like a second rank tensor not a four vector. I would suggest you look in Jackson classical Electrodynamics for the transformations. One observer will see a moving magnetic field which will have an electric field component. At low velocity this will just be interpreted as magnetic induction.


2

The equation by David Z can only be applied for particles having mass not for photons or other massless particles. The general equation is $$E^2-p^2=m^2,$$ where $E$, $p$ and $m$ are energy, momentum and mass of the particle resp. And this mass is rest mass, there is nothing like relativistic mass in modern books.


-1

Relativistic momentum is given by calculating Newtonian momentum, but instead of using mass, we use relativistic mass. Therefore, starting with $p = m_0v$ To become correct, we use relativistic mass $p = \dfrac{m_0}{\sqrt{1-\dfrac{v^2}{c^2}}}v$ Or, in a more attractive form $p = \dfrac{m_0v}{\sqrt{1-\dfrac{v^2}{c^2}}}$


9

The modern answer I know a body's momentum is equal to the product of its velocity and mass. No, it's not. At least, not in general. $p=mv$ is an approximate formula that works well for massive particles or particle-like objects traveling at slow speeds. But for fast-moving objects, or massless objects, or things that don't even count as objects (I'm ...


1

I don't know a reference but it isn't too hard to show that proper acceleration is indeed $d \phi / d\tau$. First start from the more common definition of proper acceleration which is $du/dt$. Where $u$ is the proper velocity, $t$ is the coordinate time. Applying the chain rule several times we get: $$ \frac{du}{dt} = \frac{du}{d\phi} \frac{d\phi}{d \tau} ...


0

Perhaps a better way is to redefine your energy equation to: $$ m \ c^{2} \left( \gamma_{rms} - 1 \right) = \frac{3}{2} k_{B} \ T $$ where $\gamma_{rms}$ is the Lorentz factor of the rms speed of the relativistic gas. This can be shown to be: $$ \left( V_{rms} \right)^{2} = c^2 \left[ 1 - \left( \frac{3 \ k_{B} \ T}{2 \ m \ c^{2}} + 1 \right)^{-2} \right] ...


0

There are two separate problems here that I can help you with - terminology, and math. First problem: you're not using the term "proper time"...erm...properly. The proper time of an object is the time that a clock attached to that object would read - nothing more, nothing less. So if you're wearing a watch right now, it will tell you your proper time ...


1

It looks like the answer is negative: ...


2

In special relativity particles that move faster than the speed of light are commonly called Tachyons and there is some research about the properties of such particles. Nevertheless, most physicists regards them as unphysical, because their proper time is imaginary.


0

No there is no time dilation. Both arms are part of your frame of reference.


1

If you define "length", without exception, as a constant multiple of the time between two events, it will behave no different than the time, really. Instead, try this for a definition: the time taken for light to go from $M$ to $A$ (say), with a correction for the motion of $A$ during this time, so that we get the length at the "original" time, when light ...


0

This definition works but only in the frame of the object to be measured. More generally, if you add time as a 4th dimension then a distance is defined between two events and not two points in space. That's probably why you can't generalize your definition to a moving object.


0

4-momentum transfers accordingly to Lorentz transformation, and for a photon travelling along Y axis ($\nu$ is frequency) it has such form $$p^\mu=(h\nu, 0,h\nu,0)$$ Lorentz transformation for observer moving along $x$ axis is: $$p^0\rightarrow \frac{p^0-\beta p^1}{\sqrt{1-\beta^2}},p^1\rightarrow \frac{p^1-\beta p^0}{\sqrt{1-\beta^2}}, p^2\rightarrow p^2, ...


5

That formula for momentum is only true for massive particles. Here's what is always true: A particle with a mass $m$ ($\geq 0$) can have an arbitrary momentum $p$ (in some direction, with magnitude $\geq 0$). The energy of such a particle is $$ E = \sqrt{m^2c^4 + p^2c^2}$$ The velocity of a particle is equal to $$ v = \frac{pc^2}{E} $$ When $m = 0$, $E ...


0

It would be really difficult to get experimental evidence of length contraction of an extended object. A measuring device attached to the object would contract along with it, and even if measurements were recorded and then read after the object stopped (relative to an observer's inertial frame), they would show no contraction. Thus, length contraction can ...


0

I take two mirrors parallel and facing each other. For simplicity, I'll first discuss the case that these two mirrors were and remained at rest to each other. A light pulse is being infinitely reflected between the two mirrors $M_1$, $M_2$. $E_1$, $E_2$ are the events when the pulse is reflected by $M_1$, $M_2$ [...] Accordingly (and as far as I ...



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