# Tag Info

0

For motion along a circular trajectory the acceleration changes its direction, and is not constant.

1

Vector spaces with mixed signature metrics (both pluses and minuses in the metric signature) are known and understood mathematical structures, but it's not surprising you might not have heard of them before this point: from an educational standpoint, you need everything you've learned about metric spaces with positive-definite metrics and more to make sense ...

4

I think this question is a little more low-level then the one that it's being marked as a duplicate of, so I'm going to answer it. The basic thing that you need to know is that an inner product on the 4-vector space need not have the form it has in Euclidean coordinates. That is, defining $w = ct$, it is not necessarily the case for all 4-D vector spaces ...

2

In one of my vector mechanics lectures, the lecturer said the the space time interval was the dot product of the four position vector. But then he proceeded to show it was this: $s^2$ = $\Delta r^2$ - $c^2\Delta t^2$ Where did the minus sign come from? See Einstein's Simple Derivation of the Lorentz Transformation. It's really simple. The light moves a ...

15

So what you wrote here isn't exactly what Einstein writes in the paper, and the difference there is what's causing your confusion (also he changes what he means by $\phi(v)$ halfway through the paper, which is the real problem). On page 7 of the pdf you linked, these equations appear: $$\xi = a \frac{c^2}{c^2 - v^2} x'$$ $$\eta = a \frac{c}{\sqrt{c^2 - ... 0 In order to measure the velocity in your setup, you have to measure the actual time difference for the object entering and exiting the region \Delta x, not the time difference seen by a distant observer receiving signals of the entrance and exit. There are (at least) two ways to do this. For example, station toll-takers at the locations where the object ... 0 There is a fundamental flaw in your setup. While the actual times the entrance signals may reach clocks 1 and 2 could be different, t_1'-t_1 will be the same as t_2'-t_2, because the clocks are at rest with respect to each other. In fact, consider there are two events: entrance and exit. SR says that$$c^2\Delta t^2 - \Delta x ^2 = c^2\Delta t'^2-\Delta ...

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Suppose as the object enters $\Delta$x, it hits an antenna which sends a signal to $x_1$. $x_1$ proceeds to record the time. It also does the same for $x_2$, who also records the time. A similar signal is sent when it leaves $\Delta$x. Yes the light signal will take a longer time to reach $x_2$, and thus the signals will reach $x_2$ at different times, but ...

2

The time measured by the clock you carry with you on your journey is called the proper time, and the proper time is just the length of your world line (give or take a factor of $c$). So if we calculate the length of your world line as you accelerate away then back, that gives us your elapsed time. Better still, the proper time is an invariant i.e. all ...

0

How does work function transform in Einstein's special theory of relativity? The work function of (some particular spot on the surface of) some particular piece of metal is not subject to transformation at all; instead it is a proper attribute of this piece of metal under consideration; it remains invariant under any (change of) description. A ...

5

Recall that the Faraday tensor in this form is a linear mapping that maps a charged particle's contravariant four-velocity to the latter's rate of change, wrt proper time (modulo scaling by invariant rest mass $m$ and invariant charge $q$): $$m\,\frac{\mathrm{d} v^\mu}{\mathrm{d}\tau} = q\, F^\mu{}_\nu\,v^\nu\tag{1}$$ Now let's think of a particle's ...

-3

This is tantamount to saying that the potential and internal energy of the object in a given frame is the same before and after the emission. How is this assumption justified? Because potential energy is internal energy. It isn't some other energy in some mysterious other place, it's right there in the body. To understand this you need to start with ...

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https://en.wikipedia.org/wiki/Einstein%27s_constant Start with the last part: "About constants The Einstein field equation has zero divergence. The zero divergence of the stress–energy tensor is the geometrical expression of the conservation law. So it appears constants in the Einstein equation cannot vary, otherwise this postulate would be violated. ...

0

The reason "it is not independent of the standpoint of the observer with the watch or clock, as we know from experience," is that one clock moving at one speed has just the one clock. One person gets to look at their own clock and everyone else has to defer to them to find out. But once the group is synchronized then each can look at their own clock. Now ...

1

Recall that the gradient of a scalar field is a (0,1) tensor, a one-form. With a vector field $\vec A=\vec A(\vec r)$ you can think of having three scalar fields $P(\vec r),$ $Q(\vec r),$ and $R(\vec r)$ and then $\vec A(\vec r)=P(\vec r)\hat x+Q(\vec r)\hat y +R(\vec r)\hat z.$ We will do the same kind of thing in flat spacetime. Note that the vectors ...

1

No. The charge enclosed by a surface is really just the number of protons inside minus the number of electrons inside all multiplied by the charge of the proton (which is a constant). If the electric field has a large density in one region then it has less in another so that total flux over a closed surface is still the net enclosed by the whole surface is ...

1

Entanglement experiment performed in one frame of reference guarantees that the two measured results are synchronized in this frame of reference. If we try to perform the same entangled experiment while the two ends of the same length fiber optic lines are attached to two frames of references moving with a constant relative velocity, the two measured results ...

0

Imagine a typical path. It starts at event $A$=$(ct_A,x_A,y_A)$ and ends at event $B$=$(ct_B,x_B,y_B).$ (Note I'm using $w=ct$ as the fourth dimension so we can measure everything in meters.) In the frame of the fluid this typical path has length $l_0.$ In both frames the separation is null. We can actually choose the origin of the fluid frame and your ...

1

Perhaps it's more illuminating to look at the whole thing in a spacetime diagram. we have the earth frame with coordinates $(t,x)$, and its trajectory through spacetime is the blue line. The trajectory of the spaceship is the red one. Straight worldlines are inertial frames of reference, curved or non-straight worldlines are non-inertial frames of ...

1

The axis of simultaneity, or in other words, the set of events which are simultaneous as measured in the rest frame of the ship, does indeed change suddenly when we turn back. This is because it depends on your reference frame. There isn't a single inertial frame that stays with the ship for the whole journey; you can either accept that the frame is ...

2

If you are abit more careful about making the statements, then "both perspectives" are actually correct. Let me be more concrete and explain. Let $E$ be the one who stays on Earth and $S$ be the one who is on the spaceship. The first issue you have is you said "let's say seven years passes on Earth" - this is an ambiguous statement: from whose perspective? ...

1

No. A heavier object will gain more kinetic energy when falling than a lighter object but it has a higher potential energy to start with. It is actually not kinetic energy that is needed to fall but potential energy that is converted in kinetic energy while falling. Generally, while the gravitational force on an object depends on it's gravitational mass, so ...

-3

First of all I would like to thank you for letting me participate of this discussion. I've been trying to get an answer for this question myself for at least six years now. Ever since I've realized that the experiment hasn't been done. Then one day in the same year (2009) I found a video on YouTube made by a German Blacksmith named Martin Grusenick. We've ...

0

I'm struggling to identify whether a scalar is a Lorentz-scalar. E.g: ∂iAii∈1,2,3. How do I determine if this is a Lorentz-scalar or not? If got the same problem with tensors. How do I differentiate between a tensor and a Lorentz-tensor? By "tensor" do you mean a tensor in three space dimensions? If so, then a tensor is never a Lorentz tensor. A Lorentz ...

3

Absolutely not. Both you and the people on Earth would see each other moving $7$ times slower. To repeat myself: you would see them going $7$ times slower and they would see you going $7$ times slower. This is because of the main principle of special relativity. As long as neither of you is accelerating there is nothing to choose between your frame of ...

0

You can look at the problem like this: Whether or not one person gets to see all the other persons as connected for some period of time depends on 1) the disk size, say radius R 2) the duration $\tau$ of the connection in the disk frame 3) the speed $v$ of individual components on the rim relative to the disk. Let's denote $T$ the astronauts' period of ...

-1

In theory this effect is always present, but it may be less present compared to the photoelectric effect. But it is noticeable only if the frequency of the incoming photon has (at least) a magnitude of 10^20-10^21 Hz. If the frequency used is less, the variation of the wavelength is negligible and so I could not identify it.

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The Compton effect is the inelastic scattering of photons by electrons. Compton's initial experiment used electrons in a graphite crystal to act as scatterers. These electrons are not free, they are bound, but the X-ray energies (17 keV) were large compared with the binding energies, so they approximated to free electrons. Photons of lower energies (UV to ...

0

Your confusion comes from the fact that you are IMPLICITLY reasoning using the concept "absolute simultaneity", that is: "if two events in different points of the space are simultaneous for one inertial observer then they are simultaneous to all inertial observers". One of the most important features of Special Relativity is precisely that there is NO ...

1

You have a good intuition on the possible answer because it does involve oscillations, but you need to visualize the string differently. The origin of the conundrum is not quantum, but relativistic. Here's why: Consider an inertial frame O and set up a string spanning the entire $x$ axis, stretching from $x \rightarrow -\infty$ to $x \rightarrow \infty$. ...

0

Simple answer: no. Reason: Let's say you are a photon. A mirror is placed in from of you. If you do not smash in to it then it is traveling as fast as you. there for, the photon will never reach it to be redirected backwards.

0

I am tempted to give an experimentalist's point of view, though of course I agree with Lubos and Savanah's answers. Consider a disk of radius R, then circumference is 2πR. Now, make this disk rotate at velocity of the order of c(speed of light). Here it is evident that you are defining a center of mass system for a disk and a person/machinery that will ...

0

SR is based upon two postulates: 1)The laws of physics are the same in all inertial frames of reference. 2)Speed of light $c$ is constant for all inertial frames of reference. To arrive at some results in relativity, one only needs to assume one of the postulates, other results are logical conclusions that require assuming the two postulates together. ...

2

So Special Relativity states that for all non-accelerating objects of matter the laws of physics are the same. I think the point is just that the constants and the time and space derivatives that appear in a law of physics should not have to change the form of the equation if you measure the time and the space in two frames that move relative to each ...

0

When the people see the light they fire thrusters to get up to speed and move in a circle and then after that they fire their thrusters just to maintain uniform circular motion... OK, we have a disk, and light, and a rotating ring of people. No problem. So there are frames where all of the one events happen before all of the other events. And ...

1

It is like to have two numbers each smaller than the other, it is not possible. It's not like that at all. Consider the following true statement: B observes A's clock to run slowly while A observes B's clock to run slowly. This is not a contradiction due to the crucial word observes. Consider an additional true statement: The observer at the ...

-1

To someone moving on the edge of the disk, the disk is length contracted in the direction of travel. So to them, it is not a circle, no problems with geometry. Euclidean geometry only holds for an inertial plane of simultaneity. By the end of this answer you should know when and how to use Euclidean geometry and what it means (or doesn't). And a collection ...

0

Although the answers here are good, in my opinion, they're too mathematical for the OP to get it. Before explaining what Brain Greene intended to convey in his article, I'd like to clear up your misunderstanding. There's no such thing as "time speed" vs "space speed", speed is the magnitude of one's velocity, and velocity is the rate of change of one's ...

2

This is just relativity of simultaneity again. A similar thing happens if you have a bunch of spaceships in a line that fire their thrusters at a fixed time. Different observers will disagree about whether they fired at the same time and will disagree about the spacing. Always in a consistent way. So I'd like to address the concept of geometry by not having ...

0

from the observers perspective information has travelled instantly, the firing of the first guillotine has turned on the light with no delay. Herein lies the flaw in your reasoning. The electrical signal doesn't instantly propagate, it propagates at the speed of light. From the perspective of every observer, the electrical signal from the sensor reaches ...

6

No, it doesn't violate the rules of geometry, it violates the rules of Euclidean geometry. Simple conclusion: for an observer fixed to a disk rotating uniformly relative to an inertial frame, the spatial geometry is non-Euclidean; in particular, the ratio of a circle's circumference to its diameter depends both on the circle's diameter and center position. ...

6

What is wrong is the idea that one can actually make the disk rotate; and it will remain perfectly rigid. In reality, what this correct argument shows is that relativity doesn't admit the existence of any perfectly rigid bodies. This is a perfectly basic, settled, and indisputable textbook material that every mature physicist knows. The first sentence of ...

1

Since all the velocities are constant working out $\frac{dx^{1}}{dx^{0}}$ is simple division (no calculus required). Note that the entries in your vector in $S$ are the values you're interested in: $$(dx^0, dx^1, 0, 0)=(\gamma(d {x^0}' - v d {x^1}'), \gamma (d {x^1}' - v d {x^0}'), 0, 0)$$. So we have: \frac{dx^{1}}{dx^{0}} = \frac{d ...

0

If you have two vectors you can project one onto the other. It you orthogonally project a spacetime vector onto your unit tangent then the length of that projection is how much time you observe separating the two events. That might be enough to answer your question right there. So to get an average velocity between two events first compute the vector A ...

0

Hint: $T_1=t_A+\dfrac{\text{distance covered by the light beam in the interval} [t_A,t_1] }{c}$ $T_2=t_B+\dfrac{\text{distance covered by the light beam in the interval} [t_B,t_2]}{c}$ You don't get the second terms in these two equations right. This is your only problem. Try to do it again. Your mistake was that you considered the distances traveled by ...

0

Imagine a sensor on the front of the train which detects when a guillotine passes infront of it, this then sends an electrical signal down the train to a light at the back of the train which, I think, would turn on at the same time as the back guillotine drops, from the trains perspective That's not possible. Instead of a detector on the front ...

-2

Using this equation for the events in question make the first message received at 32.732seconds Earth time and the second message 360.0595seconds Earth time. The difference between them is 327.3275seconds.

0

$T_1=\gamma*(t'_1)$ $T_2=\gamma*(t'_2)$ Above both equations are correct. But $T_1 \neq t_A+(0.4c*t_A)/c=1.4t_A$ $T_2 \neq t_B+(300*c*sec-0.4c*t_B)/c=0.6t_B+300$ Check your calculations again.

1

There are two ways to look at the error in this calculation: 1) It's the wrong Lorentz transformation. The Lorentz transformation takes the unprimed coordinates to the primed coordinates, but we need to obtain the unprimed coordinates from the primed coordinates. The correct "inverse" Lorentz transformation is: $$x = (x' + vt')/\gamma$$ $$y = y'$$  z = ...

1

This tells us that your calculations are wrong. Light from the supernova represents the information of the supernova. Information is limited to the speed of light and so the information that the supernova took place cannot outpace the light from the supernova. Even if it is along a direct line-of-sight between the supernova and Earth, the spaceship cannot ...

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