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MrBlank

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Is this correct?

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In summary, the velocity in the y-direction goes to zero because otherwise the net velocity would be greater than the speed of light. This implies that “active” acceleration applied in the positive x-direction caused “passive” acceleration in the negative y-direction.

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MrBlank

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Is this correct?

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A.T.

Science Advisor

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What about momentum conservation in the y-direction?MrBlank said:This implies that “active” acceleration applied in the positive x-direction caused “passive” acceleration in the negative y-direction.

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The answer is a resounding "it depends". Ultimately, the reason the object would accelerate is a force acting on said object and it depends on how this force is applied and in which direction. To be specific, consider a charged particle moving in a constant electric field perpendicular to the original direction of motion. Then yes, this force (which is purely in the orthogonal direction to the motion) will cause a deceleration in the original direction of motion even if no force acts in that direction.MrBlank said:

Is this correct?

It is conserved.A.T. said:What about momentum conservation in the y-direction?

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Nugatory

Mentor

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No, because velocity addition doesn‘t work the way you’re expecting. Check out the “general configuration” section of the Wikipedia article on the relativistic velocity addition formula.MrBlank said:Is this correct?

Your thought experiment is equivalent to calculating the relative velocity between two spaceships, one initially at rest and accelerating in the x direction and the other moving with constant speed ##V_y## in the y direction.

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MrBlank said:

Is this correct?

It can be correct. For instance, let the point mass be sliding with some velocity vy on the floor of "Einstein's elevator, and let the acceleration of the elevator be in the x direction. Let vx and vy be measured in some inertial frame in which vx is initial zero. Then what you describe happens with certain reasonable assumptions on the shape of the elevator floor (basically, that it's flat in the appropriate frame. This assumption about the shape of the floor actually turns out to be trickier than it looks).

We can clarify these assumptoins on the shape of the floor by simply stating that the y momentum of the mass stays constant in the inertial frame in which vx and vy are measured. But the y momentum is ##m \frac{dy}{d\tau} = \gamma \, m \,vy##, where ##\gamma = 1/\sqrt{1-vx^2/c^2 - vy^2/c^2}##

So as vx goes up, ##\gamma## goes up, and ##vy## drops to keep the y momentum constant.

Note however, that the velocity of the mass m relative to the floor remains constant, unlike the velocity vy in an inertial frame, which drops.

So there is a very reasonable interpretation of the original problem statement in which your remarks are correct, but the specification of the question is loose enough where multiple answers are possible.

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Assume you have two ships (A and B) side by side traveling in the positive y direction at Vy as measured from inertial frame S.MrBlank said:

Is this correct?

B begins to accelerate in the positive x-direction. By your argument B should lose velocity in the Y direction in order to keep its velocity as measured from frame S in order to keep its velocity with respect to frame S under c.

Which means that in frame S the line joining A and B will not be at a right angle to the Y axis.

However, in the rest frame of ship A, ship B, the line joining him to ship B always remains at a right angle to the y axis. ( as far he is concerned Ship A is at rest and it is frame S that is moving.)

This would set up a contradiction between the frames.

Instead what would happen is, that Frame S would measure ship B as always having a velocity in the x direction less than that what would make the speed of ship B less than c relative to frame S.

So for example, if the Vy is 0.8c, According to ships A and B, ship B's Vx can approach c, but according to frame S, ship B's Vx can never reach 0.6c

It is in the measurement of Vx where the effect is seen.

Really, this is not much different than the light clock example. The light pulse in the frame of the light clock bounces up and down( x axis) at c, From the frame in which the light clock has a velocity V along the y axis, the light pulse travels at a diagonal at c. The requirement that the light pulse travels at c in this frame does not require V for the light clock along the y-axis for the clock to decrease, but that they would measure the x component of the light pulse as being less than c.

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For the variant of the problem I describe, where we have a sliding block on Einstein's elevator, there are some useful formula in [[link]].

Probably the most familiar one would be the ordinary 3-velocity as a function of time, for an object with some constant y-momentum. We have to convert the z in this post to your x, as z in the link is direction of the acceleration, and the x in the link is what you call y.

Doing this conversion, the results I get are (remember, c=1):

$$\frac{dx}{dt} = \frac{gt}{\sqrt{1+g^2t^2}} \quad \frac{dy}{dt} = \frac{\beta_0}{\sqrt{1+g^2t^2}}$$

Here ##\beta_0## is the normalized velocity in the y direction at t=0.

For an explanation of the form of dx/dt, see [[this link on the relativistic rocket]] and set c=1 and a=g in the expression for d.

We can then compute the value of ##\gamma##

$$\gamma = \frac{1}{\sqrt{1-\frac{\beta_0^2 + g^2t^2}{1+g^2t^2}}} = \sqrt{ \frac {1+g^2t^2 } {1-\beta_0^2 } }$$

and that ##\gamma \frac{dy}{dt} = \frac{\beta_0}{\sqrt{1-\beta_0^2}}## is constant, thus ##\gamma m \frac{dy}{dt}##, the y-momentum, is also constant.

For the

Probably the most familiar one would be the ordinary 3-velocity as a function of time, for an object with some constant y-momentum. We have to convert the z in this post to your x, as z in the link is direction of the acceleration, and the x in the link is what you call y.

Doing this conversion, the results I get are (remember, c=1):

$$\frac{dx}{dt} = \frac{gt}{\sqrt{1+g^2t^2}} \quad \frac{dy}{dt} = \frac{\beta_0}{\sqrt{1+g^2t^2}}$$

Here ##\beta_0## is the normalized velocity in the y direction at t=0.

For an explanation of the form of dx/dt, see [[this link on the relativistic rocket]] and set c=1 and a=g in the expression for d.

We can then compute the value of ##\gamma##

$$\gamma = \frac{1}{\sqrt{1-\frac{\beta_0^2 + g^2t^2}{1+g^2t^2}}} = \sqrt{ \frac {1+g^2t^2 } {1-\beta_0^2 } }$$

and that ##\gamma \frac{dy}{dt} = \frac{\beta_0}{\sqrt{1-\beta_0^2}}## is constant, thus ##\gamma m \frac{dy}{dt}##, the y-momentum, is also constant.

For the

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Special relativity is a theory proposed by Albert Einstein that describes the relationship between space and time in the absence of gravity. In 2 dimensions, it simplifies the concept of motion and allows for easier calculations and predictions.

A point test mass is a hypothetical object used in physics to represent a particle with a negligible size and shape. It is often used in calculations involving gravity and other forces.

In special relativity, acceleration is defined as a change in velocity over time. In 2 dimensions, this can be visualized as a change in the direction or magnitude of an object's velocity on a 2-dimensional plane.

The key equations for calculating acceleration in special relativity in 2 dimensions are the Lorentz transformations and the equations for velocity and acceleration in 2 dimensions. These equations take into account the effects of time dilation and length contraction.

Special relativity in 2 dimensions has many practical applications, including in the fields of particle physics, cosmology, and GPS technology. It also plays a crucial role in understanding the behavior of objects moving at high speeds, such as spacecraft and particles in accelerators.

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