# Tag Info

-1

I like to think of it this way, say you have a mirror like sphere (such as a silicone bulb you would find on a bird bath) now you look at the your reflection and it looks fairly normal, THAT CAN REPRESENT YOUR REFERNCE FRAME, and then you look at the edges of the bulb, THAT CAN REPRESENT THE SPEED OF LIGHT BOUNDARY, but you notice that any reflection near ...

1

why wouldn't that straight line be in the direction of acceleration why do you think the acceleration line be in the direction of tangent? the tangent is where a body would have kept moving if the rope didn't pull it. so the vector of speed changes towards... where the rope is attached, i.e. perpendicularly. acceleration is the change in velocity ...

2

Similar questions have cropped up on this site many times, and the debate surrounding them is usually fractious because people misunderstand each other's use of words like exist. One of the lessons of General Relativity is that any observer has to choose a locally convenient coordinate system that may not be globally convenient. We on Earth (quite sensibly) ...

1

It's a matter of what you mean by "see". Even for a distant observer, it will take a small amount of time for the gravitational redshift effect to become essentially infinite. If your collapsing gas star redshifts to the point where it won't emit a single photon in the age of the universe, it may not have yet technically "redshifted to zero", but it has ...

0

Assuming the electrons are moving at the same velocity the beam would look the same because the relative velocity is zero between the electrons. However, according to the scientist(lab frame), the beam gets shrunk (and according to the electrons the scientist gets shrunk).

0

Remember that you must always specify the inertial frame of the observer. Other than that, your question makes perfect sense. The "closing" velocity of the two photons approaching each other will be 2c only in an inertial (stationary) observer at rest relative to the center point. To an observer "riding" with one of the photons, (either one), the closing ...

0

You are quite right. Energy is not conserved between the reference frames. That is the biggest mystery. It is certainly going to change one's concept of understanding of energy. Here is an example where there is total failure of 'law of conservation of energy' Take an example of spaceship in space. suppose you start the spaceship and accelerate for one ...

0

The world lines of both clocks pass through two particular events ('points' in spacetime), the event of your leaving the home and the event of your returning. The worldline of the clock at home is straight while the worldline of the clock in your pocket must be curved due to the acceleration you undergo during your near light speed trip out and back. A ...

0

Let's say we have two clocks. Let's call the (coordinate) times of these two clocks $t_A$, and $t_B$, respectively. I leave one at home and keep one in my pocket. Then, I started running [...] then come back to my house. If I compare those two clocks how would they differ in time? If it is also given (corresponding to the comment by the OP ...

0

Technically they wouldn't differ in time because to come back to your house you would have to decelerate and accelerate, which is under the realms of general relativity or as seen in comments below (thanks to @dgh) we can use comoving frames. You could use two frames one moving towards school and one away from school and Lorentz transform between the two, or ...

0

The question of what is the velocity of a photon relative to another photon does not make sense. Neither it does asking what is the velocity of anything relative to a photon. This is because in special relativity we only have the concept of a velocity defined for a massive observer, which is defined from the four-velocity $$u^\mu = \frac{d x^\mu}{d\tau}$$ ...

1

What is angular speed? Clearly it is $\frac {v_\perp} {r}$ where symbols have their usual meanings. Rod rotates about its, say, rightmost point, say $O$. We will consider left side as positive $x$-axis. Now consider a point $A$ at distance $r_1$ from it. Let the rod have instantaneous angular speed $\omega$. All points on the rod will have this $\omega$ ...

0

Anupam has very well answered, I'll just add my modest contribution with a simple example: the inertial frame could be the Earth and the non-inertial frame a car moving in a circle. In that case the centripetal force is the reaction of the tyres on the ground due to their grip, but the most interesting thing about this case is that objects in the car seem to ...

6

A centripetal force is not a fundamental force. We call any force a centripetal force if it is acting towards the center of the direction of rotation, perpendicular to the direction of motion. Rotating a rock tied on a string? Centripetal force = tension in the string Satellite orbiting Earth? Centripetal force = gravity Charged object rotating around an ...

1

The actual question in this question, is a good physics question. Freely interpreted, it basically asks if SR effects, in particular time-ordering of spacelike separated events, make it difficult or impossible to simulate physics. The answer to that is no. An "external" Simulator (be it a particle physicist or the hypothetical people simulating our ...

1

The important thing is that all speeds seem to change: If you would look at the universe, in this case the two protons, from the perception of a man shrunk to the size of a proton, not only would the particles appear much faster, but so would the speed of light. So if you shrunk yourself to a trillionth of what you are now, one proton would have a diameter ...

3

If I understand you correctly, your two points about apparent slowness of speeds is related to scale, and disappears when you quantify it using a common unit. ie: We think of 10m/s as relatively slow because the average human is 1.8 metres in height, and we can imagine that 10 metres per second, or 36 kilometer/hour as an achievable speed using a machine ...

2

The use of inertial frames in Lagrangian mechanics is by no means compulsory and everything can be done in any reference frame provided one takes all forces, real and inertial, into account. Actually there are two possibilities in interpreting the question. We work in a non inertial frame $R'$ (instead of an inertial one $R$) because we are adopting ...

3

It depends on how you "derive" Lagrange's equations, whether taking Newton's laws as fundamental or by assuming an action integral and minimizing it. However, there is no such requirement that you be in an inertial frame of reference. Thus, to look at your pendulum problem, you could start with the Lagrangian L = \frac{1}{2} I ...

3

I) In e.g. Ref. 1 is shown that there exist (possibly velocity-dependent) generalized potentials for all the fictitious forces, such as, e.g., the centrifugal force, the Coriolis force and the Euler force. So Yes, there exist Lagrangian formulations for non-inertial accelerated reference frames. II) OP's image shows Kapitza's pendulum. Kapitza's pendulum ...

-1

No. It would not be possible. Things get very hot around the edge of a black hole. There is no technology that exists or is likely to exist that could protect a camera. The fact that EM radiation is in fact light and the strength of gravity...lalala. No. It's just not possible.

1

No, not even light can escape a black hole, therefore radio will be unable to broadcast signals back to Earth. (Radio is a form of light)

2

Actually the COM for the 2-body problem is the essential feature in this subject and with respect to it, both the Earth and the Sun rotate. Indeed, motion is relative, the relativity of it is even easier to understand in the Galilean Relativity than in Special Relativity. The Heliocentric view is actually the correct opinion that the Sun of our planetary ...

0

I'm going to take a slightly different approach to explaining this, in analogy with a great answer about the ontological nature of Newton's Laws. First, let's posit the existence of an inertial reference frame. It doesn't matter which one, but there has to be one. This is an important point, and one that's often overlooked. In it, nothing is moving faster ...

0

Both observers would perceive the light at the same speed. Any observer in any frame of reference anywhere in the universe will see light travel at the same constant speed. In terms of an light under gravity - if the light is approaching the massive object it will have a Doppler blueshift, while if the light is going away from the massive object it will have ...

3

General relativity reduces to special relativity locally. What this means is that given an error tolerance $\varepsilon$, you can find an extended region (perhaps just a small one) around any point in spacetime such that the laws of physics as tested only within that region match those of special relativity to within $\varepsilon$. That means if you measure ...

1

You have to measure everything from the same frame of reference. Your own frame of reference obviously has a velocity of 0, relative to you. The other object moving toward you, or away from you, will never move faster than the speed of light as seen from your frame of reference. A third observer can see two objects, each moving at the speed of light. Toward ...

1

This is the reduced mass. This effectively accounts for the fact that the center of the sun is not 'fixed', but rotates around the center of mass (also called the barycenter) of the Earth and the sun together. The expression $\frac{mM}{m+M}$ approximates to $m$ when $M>>m$.

Top 50 recent answers are included