24
votes
Accepted
Wikipedia states that the relativistic Doppler effect is the same whether it is the source or the receiver that is stationary. Can this be true?
You have to include a factor of $\gamma$ in both effects. The terms that are the same are $\gamma (1-\frac v c)$ and $\frac 1 {\gamma \left(1 + \frac v c \right)}$. This is because
$\displaystyle \...
16
votes
Wikipedia states that the relativistic Doppler effect is the same whether it is the source or the receiver that is stationary. Can this be true?
The relativistic Doppler factor has to be the same whether it is the receiver or the source that is considered to be stationary, because in relativity there is no way to determine if it is the ...
6
votes
Wikipedia states that the relativistic Doppler effect is the same whether it is the source or the receiver that is stationary. Can this be true?
But your final two expressions are approximately the same in the limit $v\ll c$. Use the binomial expansion,
$$
\begin{align}
(1+\epsilon)^n&= 1+n\epsilon+\frac{n(n-1)}{2!}\epsilon^2+\cdots
\\&...
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4
votes
A puzzle about relativistic spin
With all thanks to Valter Moretti, I believe we now have a complete solution to my original puzzle. In the identities for $\mathbf{K}$ and $\mathbf{J}$ that disagreed with the celebrated ones due to ...
4
votes
is $LT$ a Lorentz invariant?
Consulting any text-book on special relativity, you will find the well-known fact that the four-dimensional volume element $d^4x = c \, dt \, dx \, dy \, dz$ is a Lorentz invariant as a consequence of ...
4
votes
Accepted
Is there a situation in which two photons/particles violate Lorentz transformation?
What you are not taking into account is the relativisic velocity addition.
When calculating the gamma factor we are calculating it for a single reference frame moving relative to our reference frame. ...
4
votes
Accepted
Some confusion about understanding the relativistic quantum mechanics
The group is still the Poincare group (Lorentz+ translations). The tricky thing is that we need to find a way for that group to act on the Rays in such a way that it preserves the probability. There ...
3
votes
Meaning of general Lorentz transformations
Let's look at the bigger picture first. A Lorentz transformation maps a set of coordinates $(t,\mathbf r)$ to another set of coordinates $(t',\mathbf r')$. More specifically, if an observer $\mathcal ...
2
votes
Meaning of general Lorentz transformations
I will call these observers $S$ and $S'$. The condition $\mathbf{r}_{\parallel}=\mathbf{v}t$ implies that $S$ is measuring a point in space that travels with velocity $\mathbf{v}$, with the same ...
2
votes
Wikipedia states that the relativistic Doppler effect is the same whether it is the source or the receiver that is stationary. Can this be true?
For sound waves propagating in a medium, there is an obvious absolute frame of in which the sound wave exists with a frequency that does not depend on the motion of the source (Tx) or receiver (Rx).
...
2
votes
Accepted
Resolution of Ehrenfest paradox using only special relativity
Instead of a solid disk, we might as well think of a circular train traveling on a circular track.
Alice sits in the station. Bob is riding on the train.
The Question: The track is shorter in Bob's (...
1
vote
Is there a situation in which two photons/particles violate Lorentz transformation?
What you did was introduce a third inertial frame in which the relative velocity between the two particles is indeed higher than $c$. This does not violate Special Relativity. If you go in the ...
1
vote
When is the Lagrangian a Lorentz scalar?
An obvious, kind of dumb, answer is that the Lagrangian corresponding to a given Hamiltonian will be a Lorentz scalar if the Hamiltonian has the form,
$$ \mathcal{H} = \pi^a \frac{\partial}{\partial t}...
1
vote
When is the Lagrangian a Lorentz scalar?
As far as I know, there are no good ways of stating what conditions on the Hamiltonian will cause the Lagrangian to be a Lorentz scalar other than to just say the Hamiltonian must be derived from a ...
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