# Tag Info

## Hot answers tagged special-relativity

55

For simplicity, consider the case $u=v$. The "slow" formula is then $2u$ and the "fast" formula is $\frac{2u}{1+(u/c)^2}$. In the plot you can see these results in units of $c$. The "slow" formula (red/dashed) is always wrong for $u\ne0$, but it is good enough [close enough to the "fast" formula (blue/solid)] for small $u/c$. The cutoff you choose depends on ...

28

I'm usually a little more specific with my students. Let's consider a car traveling down the highway at $\rm 30\,m/s \approx 60\,mph$. Then the denominator in the relativistic formula is something like $$1 + \left(\frac vc\right)^2 = 1 + \left( \frac{30\rm\,m/s}{3\times10^8\rm\,m/s} \right)^2 = 1 + 10^{-14}$$ In your car, then, the difference between ...

9

Actually, regardless of the velocity of the objects in concern, the 'velocity addition' formula is always $\frac{u+v}{1+uv/c^2}$. There is no transition point where the formula changes from $u+v$ to the special relativity one. Its just that the difference you get in both the formulas at 'low velocities' is very very negligible. $c^2$, in $m^2/s^2$, is ...

5

A force is defined as a change in momentum over time. In the Newtonian limit, this means a mass times an acceleration. But when dealing with things like photons, the formal definition is applied. Photons have no mass, only momentum. Therefore, if a force is applied to them, their momentum can be changed. This can happen in two important ways, a force can ...

4

Build up four-momentum from the mass and the four-velocity: $$m \vec u = m \gamma (\vec e_t + \vec v)$$ But four-momentum can always be decomposed according to energy and momentum: $$m \vec u = E \vec e_t + \vec p$$ Equate the $t$-components. Done.

4

Because the photon definitely has energy, it must have a four-momentum vector, but it must be defined differently from mU because the proper time, $\tau$, along its worldline is zero. $$d\tau= dt\sqrt{1-v^2/c^2}$$ The photon four-momentum vector is defined to be ...

3

The word "relativistic" means "compatible with principles of Special Relativity". This usually implies that we can no longer use the "classical" picture of universal stationary space and time. Instead we talk about 4-dimensional space-time. The word "quantum" means compatible with principles of quantum mechanics. You can look them up on wikipedia etc. But ...

3

Well, you could treat them separatedly, via two equations, say $$\mathbf{F}_\text{elec}=q\mathbf{E}$$ $$\mathbf{F}_\text{mag}=q\mathbf{v}\times\mathbf{B}$$ but since Newton's second law holds, in presence of an electric field and a magnetic field, the total force will be the sum of both, that is, the Lorentz force. I would say is just as simply as that.

3

In a rotating reference frame, the coordinate velocity of an object can exceed $c$. However, this doesn't mean that they're moving "faster than light". If we were to look at the light-cones at these distant locations, we would see that the four-velocities of these objects are still confined within the light-cones at those locations. To put this another ...

2

The "fast" formula is always the correct one. A "fast" speed is one that is comparable to the speed of light. However, when both the speeds involved are much smaller than the speed of light, the "slow" formula is a very good approximation.

2

Energy is not a Lorentz invariant quantity, it is the zero-th component of the four-vector. Only proper orthochronous Lorentz transformations preserve the sign of the zeroth component, so if the energy is positive in one frame, a non-orthochronous Lorentz transformation would yield a frame in which the energy is negative. But we usually only allow the ...

2

The photon is an elementary particle. There are two ways do measure the frequency and therefore the energy of the photon since its energy E=h*nu . 1) using a diffraction grating which analyses the wavelengths in a beam of light , as below: This is the spectrum of iron. Each line is composed of zillions of photons with that frequency. If one sent one ...

1

In a comment Kyle points out the question Does strong magnetic field cause time dilation?, which is closely related to your question. However it isn't a duplicate because you are specifically asking whether experiments have been done, not whether the effect theoretically exists. Theoretically we would expect a magnetic fied to contribute to the curvature of ...

1

Assuming tentatively that all velocities and notions of simultaneity are in the frame of reference of A, i.e., the events C/D passes A/B occur simultaneously with the event B reaches Z in the frame of reference of A, and C/D are moving past A/B at speeds 0.6c as measured in the frame of reference of A. Then the time will be offset by 2 years. If A and B ...

1

When does a object go fast [enough to require the use of $\frac{u+v}{1+ uv/c^2}$]? When you are approaching light speed. How fast do you need to go to be "approaching" light speed? That depends on the precision. The key is that $\frac{u+v}{1+ uv/c^2}$ always works. However, it's not exactly simple, so generally we like to use an approximation: ...

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