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3

The radiation emitted by an accelerated charge depends on the boundary conditions on the fields at infinity. When one takes this into account properly, then accelerated observers will agree with inertial observers about the emitted radiation (after trivial transforms are applied). Any treatment which purports to show that in the accelerated observer's frame ...


2

Assuming that Classical Electrodynamics (Maxwell's Equations) holds, the answer is that the inertial observer would see the radiation while the non-inertial observer would NOT. The question you are asking is basically the following paradox: https://en.wikipedia.org/wiki/Paradox_of_a_charge_in_a_gravitational_field This paradox has been analyzed and ...


1

It depends on whether the current is carried by a conductor or is in free space (an electron beam). In the case of an electron beam, the current will appear to have reversed in direction if you travel faster than the charge carriers, even without relativistic effects. This web page does the transformation roughly like you have attempted, using a charge ...


2

Also, the fact that $\Delta t' \rightarrow 0$ if we formally let $v \rightarrow c$ can be interpreted as saying that no time at all passes for a particle moving at the speed of light. Photons cannot "age" or in any other way change over time.


4

Yes: I've done that. I used to have a device for the purpose, commonly called "rollers". It's like a treadmill for bicycles.


-1

I would say it is the tire deformation keeps up the bike rather than unconscious steering tweaking; at high speed, high frequency tweaking is impossible. Imagine or if you see the cross section of a tire, you can see it is least deformed when the portion is not close to the ground. And it is flattening deformed when it is in contact with ground. The ...


-1

Roughly speaking, a bike doesn't tip over because as soon as you start to tip over, gyroscopic effects (those weird forces you feel when you try to rotate a fast spinning object) together with the structure of the bike cause it steer towards the direction of tipping. Turning in the direction of tipping causes a centrifugal force which will turn the bike ...


28

What keeps a bicycle up is a variety of things, but it all comes down to the front wheel, which can move left/right. The bike is always out of balance, and if it starts to fall to the left you unconsciously turn to the left, which moves the point of support (the wheel on the surface) to the left, which arrests the fall and may start the bike falling to the ...


19

By the principle of relativity, you will not fall over – assuming that you know how to use the bike and you won't be deliberately "confused". The principle says that the laws of physics have the same form in all inertial frames that are moving by a constant velocity relatively to each other. The reference frame associated with the moving sidewalk is as good ...


1

It's the time coordinate of an event in the $S$ frame (the coordinate frame you're moving with respect to), then a fixed time. Important remark: the equation that you are using does not give the decrease (not inflation) in length of an object as measured by an observer who's moving with respect to the $S$ frame, but the $x$ coordinate transformation between ...


3

Here's a derivation that uses very basic properties of space and time (isotropy, homogeneity, the fact that two Lorentz boosts should compose into another valid Lorentz boost, etc.). The constant maximum speed through space (i.e., the speed of light) is a derived property, not an assumption. One more derivation of the Lorentz transformation - Jean-Marc ...


1

Science does not say that the speed of light is constant. Rather, a lot of people with fancy clocks and tape measures have expended a lot of effort measuring the speed of light and have discovered, to their initial surprise, that it is constant. Then another bunch of people (largely one person) came up with a model - a bit of mathematics, otherwise known ...


2

You have this backwards. What you state, the fact that "time is relative", is a consequence of the theory rather than a cause. Assume some light ray travels some distance in reference frame 1 and is seen also from reference frame 2. The trajectory is obviously different in both frames. That's nothing special that Einstein would have introduced : just try ...


2

Yes, I agree with your interpretation of Minkowski's statement. You might be interested in reading the answers to What is time, does it flow, and if so what defines its direction? as I discuss exactly this point in the first part of my answer. The key point is that there is no observer independent way of separating the time and spatial coordinates so ...


2

I use other symbols in order to prevent confusion in the following. Let a point charge $\:q\:$ moving with position vector $\:\boldsymbol{\xi}\left(t\right)\:$ as in above Figure. Then the volume charge density and the charge current density are expressed via Dirac $\:\delta$-function as follows \begin{align} \rho\left(\mathbf{x},t\right) & ...


2

As per, http://en.wikipedia.org/wiki/Four-velocity, we can define four-current density as: $J = \rho_0 U$, where $U$ is the four-velocity. Since it's a scalar times a four-vector, it's another four-vector. $$J = \gamma(v)(\rho_0 c,\rho_0 \vec{v})$$ $$J = (\gamma(v)\rho_0 c,\gamma(v)\rho_0 \vec{v})$$ Now it remains to show that this fits the definition you ...


1

You are missing what Landau wrote before that part: "If we were to choose an arbitrary frame of reference, space would be inhomogeneous and anisotropic". If you choose an accelerating frame of reference, fictitous forces will be present and free bodies will start to move even if no force is acting on them. As Landau writes, you have to choose a frame in ...


2

You caught that the barn doors do not close and open at the same time in the vaulter's reference frame, but missed the issue that they are moving in her frame. So, the difference in position between where they were when they flashed shut will not be L/4, but L/4 + vΔt. Include that, and you should be able to solve it.


3

The basic idea is that physical laws are same in all inertial frames. Framing your question in a different way: Why do we generalize a formula(which gives time-dilation) whose derivation is based on a light clock to physical clocks and even the biological clock? A very interesting argument was given by Feynman in his Lectures on Physics, Vol:1. ""To ...


0

The special theory of relativity is a new theory of the space and time and all phenomena in the spacetime. The key constant that determines how strong the new "quirky" effects are is the speed $$ c = 299,792,458\,{\rm m/s} $$ which is the speed by which the "spatial directions" may be converted to the "temporal direction" in the spacetime. Whenever the ...


3

Often, I think a nice way to untangle the mess we find ourselves in is by appealing to the all-knowing God of special relativity, blessed be her name, Lorentztransformalia. Let event $A$ denote the front door closing, and let event $B$ denote the rear door closing. Without loss of generality, we assume that we've chosen our coordinates so that in the ...


1

The object doesn't move relative to fixed space--there is no fixed space per special relativity--so forget that. The question is "why is there frame dragging?": because there is normal Newtonian-like gravitational attraction, and you need the frame-dragging field to make the whole thing relativistically invariant--that is, consistent for all inertial ...


0

In this nice reference the autor assumes the relativity principle + homogeneity + isotropy and deduce the general coordinate transformations which contains both Lorentz and Galileo transformations. Further he imposes the postulate of the constancy of the speed of light, restricting the transformations to be the Lorentz type.


0

In a sense, the two definitions you mention are the same. One of the postulates of special relativity is that the speed of light is the same in all reference frames, so the definition of "the coordinate transformation according to the postulates of special relativity" is the same definition as "the coordinate transformation under which the speed of light is ...


2

There are lots of ways of approaching special relativity. My own preferred approach is the invariance of the line element. Suppose you move a small distance in spacetime $(dt, dx, dy, dz)$ then the length of the line element $ds$ is defined by: $$ ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2 \tag{1} $$ This equation is known as the metric equation and is derived ...


0

In the quote cited you could imagine that point particles move in a straight line at a steady velocity and don't rotate when far from massive objects. So to a small accelerating frame, they look like a non rotating point particle moving at a constant velocity would look to an accelerating frame. But maybe point particles near a massive object have similar ...


0

Newtons 2nd law only holds in an inertial frame. If you use the center of mass of the rolling object as your frame of reference then the frame will be accelerating and F=MA won't hold. However if you fix the frame of reference to a relatively stationary point, say the surface of the earth, it will. Even in this case the frame is not truly inertial, but ...


2

Hint for (a) : Use the Figure above to prove that : \begin{equation} \tan\left(\phi^{\prime}-\phi\right)=\dfrac{w\sin\phi}{c-w\cos\phi}=\dfrac{w\sin\theta}{c+w\cos\theta}\;=\;-\tan\left(\theta^{\prime}-\theta\right) \tag{a-01} \end{equation} This explains why I find the last equality instead of the first as I post in one of my comments : we refer to ...


0

I have to say I found Newton's laws very uneasy to understand when I was in high school. I had a lot of questions similar to the OP's. I remember I asked my high school teacher exactly the same question in the OP's post. I was also confused by the meaning of mass, whether the second law is a law or a definition, etc.. My high school teacher couldn't answer ...


2

a) Assume that light has to travel 1 m to reach the origin of the reference frame I, so it'll take $t=1/c $ seconds to reach. Now, using trigonometry (the sine rule in the first equation), we get: $$ \frac{\sin{\Delta\theta}}{\omega t} = \frac{\sin{\theta}}{ct} $$ $$ct\cos{\Delta\theta} = 1-{\omega t}\cos{\theta}$$ Dividing the first equation with the ...


0

From those we can get $f(x(t),t) = x + Ct + D\,$ after integration. How can I interpret this $f(x(t),t)$? The constants of integration $D$ and $C$ are, respectively, the displacement between the origins of the two frames at time $t=0$ and the constant velocity with which the origin of the primed frame moves with respect to the origin of the unprimed ...



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