Hot answers tagged reference-frame
10
This is just a footnote to Crazy Buddy's answer (which is correct! :-):
Length contraction is a real phenomenon, and indeed the RHIC observes this every day because the nuclei are moving so fast that the collision is between two disks not two spheres.
However to see something you need to have light emitted from the object reach your eye, and the light from ...
10
Your question is a natural one to ask, but it has no answer. It is a bit like asking by what mechanism the angles of a triangle always wind up adding to 180 degrees (in Euclidean geometry). There is no mechanism for that - no one is going around checking all the triangles to make sure their angles add up right. It is just a logical consequence of the theory ...
9
You have successfully discovered that the kinetic energy depends on the reference frame.
That is actually true. What is amazing, however, is that the fact that kinetic energy is conserved is NOT reference frame-dependent. So, when you balance your conservation of energy equation in the two frames, you'll find different numbers for the total energy, but ...
8
Both are right. Any moving clock is slower than a clock at rest, from the perspective of the frame at rest.
Maybe this simplified freehand graphic (apologies for its lack of precision) helps to see that both A and B feel the same about each other's time dilation:
Let's say that the red axis represents A and its proper time measured in minutes (first ...
8
The sphere is contracted in the horizontal axis and perceived as an ellipsoid. This is what we believe about length contraction and this happens only, when we take Einstein's simultaneity into account.
But, the stationary observer would see the sphere appearing as the sphere always (i.e) the circular outline would still be there at any velocity relative to ...
6
You can't travel at the speed of light. So it's a meaningless question.
The reason some people will say that time freezes at the speed of light is that it's possible to take two points on any path going through spacetime at less than the speed of light and calculate the amount of time that a particle would experience as it travels between those points along ...
6
(I will assume in my answer that people have read the discussion on the old question, linked to by the OP.)
No, it is not like the aether. It is still true that locally, there is no preferred reference frame. You don't even really need to think about spacetime to see what is going on. Consider a two-dimensional plane, parametrised by $(x,y)$, and roll it ...
5
To elaborate on Mark M's answer:
If you consider an accelerating reference frame with respect to Rindler coordinates (where time is measured by idealized point-particle accelerating clocks, and objects at different locations accelerate at different rates in order to preserve proper lengths in the momentarily comoving reference frames), then light may not ...
5
Objects, defined as things with mass, don't move at the speed of light. The time dilation factor is
$$\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$$
and it has no limit - it diverges at $v\to c$. For speeds very close to the speed of light, we could define $\epsilon = \frac{c - v}{c}$, then we'd have $\gamma \sim \frac{1}{\sqrt{2\epsilon}}$ This shows how much ...
5
From the comments to user16307's answer I'm guessing you're fairly new to special relativity. Until you get familiar with the subject it's very dangerous to throw around concepts like time dilation and length contraction because you can easily fall into traps like the pole in a barn paradox. The only safe way to work out what happens is to use the Lorentz ...
4
In relativistic mechanics, there is a conserved quantity, relativistic momentum:
$\vec p = \gamma m \vec v$
$\gamma = \dfrac{1}{\sqrt{1-\frac{v^2}{c^2}}}$
where m is the invariant mass or less precisely, the rest mass.
Now, one interpretation is to identify $\gamma m$ as the relativistic mass, a speed dependent mass. But this is actually unnatural as it ...
4
The three-torus geometry or, more precisely, the $R\times T^3$ spacetime geometry, breaks the Lorentz invariance to nothing. The identification $x\sim x+2\pi R_x$ and similarly for $y,z$ needed to define the torus is an identification that doesn't stay the same when $t,x$ are mixed by the Lorentz transformation, so the spacetime is just not Lorentz-symmetric ...
4
in relation to anything else that can make such measurements.
As the speed of light is universal, nothing can see any other massive field moving at the speed of light (which is reserved for massless fields)
your 0.51 number suggests that you expect that naive addition of velocities holds when velocities approach the speed of light. This is wrong. Here is ...
4
There can be at least two different flavors of paradoxes. In one, a result such as 2+2=5 is proved, and the problem must be either incorrect reasoning or a set of assumptions that was invalid. In the other type, exemplified by the EPR paradox, the correct result of an argument is so surprising that it seems like it must be a mistake.
Based on the ...
3
As dmckee comments, this questions is 'wrong' since the massive trains can't go at the speed of light.
But we can modify the question a bit to say that our 'trains' are massless. And for simplicity, let's make these 'trains' massless point-particles A and B.
According to relativity, the plane of simultaneous events of train A is light-like, and also ...
3
Does this simply mean that any sound theory expressed in K, should be able to withstand a transfer to another system Z and still hold "true"? Or is there more to "uniform translation relatively to K"?
The 'uniform translation ...' part is crucial. "K' moving in uniform translation relatively to K" means that the relative velocity between them is ...
3
You have two facts.
The observer standing on Earth sees the light ray as taking eight minutes to cover the distance from the Sun.
The ray itself sees the universe as infinitely compressed so that no time elapses on it's travels (over any distance).
Both of the facts are correct. Both frames are equally valid, and physics works in both (though the zero ...
3
To expand on Qmechanic's point, we can demonstrate how null directions get moved around even in flat space by Lorentz transformations:
Given a Lorentz vector $X^a$ you can construct a 2x2 Hermitian complex matrix $$X^{AA'} = \frac{1}{\sqrt{2}}\left(\begin{array}{cc}X^0+X^3 & X^1+iX^2 \\ X^1-iX^2 & X^0-X^3\end{array}\right)$$ The Lorentz norm of ...
3
No.
Your mass doesn't increase when you move at relativistic speeds. Nothing changes at all. You feel exactly the same. Although the world around you will appear distorted, nothing endogenous to your spaceship feels any difference. This is in fact the fundamental idea of relativity. See this Wikipedia article for more detail. Also check out Galileo's ...
3
First, your wave equation is wrong. You can see this from dimensional analysis. It should be
\begin{align}
\frac{\partial^2 \phi}{\partial t^2} = c^2 \frac{\partial^2 \phi}{\partial x^2}
\end{align}
Second, you made a mistake in the cross terms for the $\partial^2 /\partial x^2$ term. The cross term should have the coefficient $-2\gamma^2 v/c^2$.
Third, ...
3
Your logic is correct but for one misunderstanding right at the beginning. The observer is smart enough to know that just because the photons are received simultaneously doesn't mean they were emitted at the same time. The emitting of a photon and the receiving at a different time and different place are two distinct events, as explored further in this post, ...
3
For both interpretations, the answer is 'yes' since force still acts in an opposite force on anything which has mass. As you accelerate, your velocity increases and therefore mass will increase. The increase in mass will bring about an opposite force. The greater the mass, the greater the inertia.
2
It's not obvious what you mean by "a time dilation effect infinite in both directions". Did you mean into the past as well as into the future?
A massless particle experiences no flow of time; not into the future and not from the past. Just to make things even stranger it experiences no distance either i.e. as far as a photon is concerned there is infinite ...
2
but I find it funny that a theory tells me there are spacetime events where I can never get to (outside the light cone).At the very least, I feel this is somewhat redundant - why not drop it?
Mathematical theories do not come a la cart, i.e. they are not patched together, cut as you go, constructs. Theories are axiomatic self consistent and sustained. ...
2
Yes: if a theory is sound then its physical predictions of phenomena will be the same regardless of which frame the analysis is done in. Note, though, that some intermediate stages of the analysis, such as electric and magnetic field, may look different in different frames, but the final result is always the same.
This presupposes, of course, the fact that ...
2
Yes, exactly. However, Albert Einstein beat you to this discovery by about 100 years with the equivalence principle. The key idea is the equivalence between a downward gravitational acceleration and downward force due to an acceleration upward. There is no experiment you can locally perform that will tell you whether you feel heavier because the elevator ...
2
My initial reasoning was wrong, sorry, I can't justify the use of angular momentum conservation in that non-inertial frame. One can write a balance equation for the angular momentum, but after all it should be the same as solution №2.
But ok, let's choose a single inertial frame. Let it be the one moving with the ultimate ball velocity $v_f$. This will ...
2
There are fictitious forces when you define a non-inertial frame.
a1: One corollary of Newton's third law is that an object cannot exert a force on itself. Another corollary is that all forces in the Universe have corresponding reactions. The only exceptions to this rule are the fictitious forces which arise in non-inertial reference frames (e.g., the ...
2
From inside the room I would view the light moving north at the speed of light
That's true.
From outside the room, observing from a stationary position, would the light exiting the window move at twice the speed of light? or would it change speed?
No, it will still move at speed of light, thus theoretically outer observer will never see any light ...
2
You're quite correct, you'd write the opposite velocity with a negative sign. You just need to decide what sign convention to use. In your example you're only considering motion in one dimension, so you could take motion to the right to be positive in which case motion to the left would be negative. Or you could take motion to the left positive and motion to ...
Only top voted, non community-wiki answers of a minimum length are eligible
