Physics 62 Special Relativity (6 of 43) Relativistic Velocity: Another Example

[@MichelvanBiezen]

We are in the process of making some more special relativity videos. They should be uploaded in a few more weeks.

[@Xxac1348xX]

Amazing explanation, my textbook and my professor lacked in explaining this, but now I understand the Lorentz transformations.

[@huynguyenanh8984]

Thank you so much for your clear explanation, I've been struggling to identify these 3 quantities for a long time!

[@MichelvanBiezen]

Glad it was helpful! 🙂

[@charifabderrahmane9105]

This is what we call a life saver.

[@MichelvanBiezen]

Glad it helped. 🙂

[@函頤]

I love your videos sooo much ! They just saved my midterm examination... (I'm a student from Taiwan

[@michelle-jh2uf]

thank you for making education videos like these you really make learning enjoyable and you sound enthusiastic when teaching! also your bowtie is really cute thank you!

[@MichelvanBiezen]

Glad you like them! Thanks for the comment.

[@-_Nuke_-]

The problem of this demonstration is that V and U are written in such a way that is hard to distinguish which is which... This might sound like a non importand issue, but for someone who has never done these calculations before its making him a GREAT disservice. I wish better terminology was used so that its clear which is which because U and V can look so much similar to eachother!

[@MichelvanBiezen]

I agree with your comment. Unfortunately that is the standard method of notation. As a student this bothered me as well especially since u was typically not used to express velocity. But now that I have seen it often, it becomes second nature and that is probably the best way to look at it. Familiarity is the key.

[@chavab8753]

Two questions: 1) Often we assume each observer views himself as stationary. When does he get to "know' he is moving? 2) What would this look like to earth observer? I think it would be the same answer: they are approaching each other at 0.946 c. Is that right? What is the nomenclature to describe this? Thanks in advance.

[@MichelvanBiezen]

The reference is motion relative to space. Earth is mostly stationary relative to space (the motions of the Earth are very slow relative to c) and thus it is easy to tell which frame is moving and which frame is stationary.

[@-_Nuke_-]

"When does he get to "know he is moving?" You don't get "to know" that you are moving or not. You are allowed to ASSUME that you are not moving and thus your can CONSIDER yourself to be at rest.

[@Official-tk3nc]

Sir i am a very big fan of you teaching. You are my favourite teacher.

[@simongarcia2336]

I somehow always end up watching your videos (and understanding it very well too!) thanks!

[@MichelvanBiezen]

Happy to hear that!

[@sambichagwa1119]

omg taking this class and this in particular confused me thank you so much it is by far clearer

[@MichelvanBiezen]

Glad we were able to help.

[@Abhijitdas8710]

Amazing explanation sir....Very helpful

[@Abhijitdas8710]

Sir i have a question??Plz help me to slove this prblm Two Spaceships A & B of proper length 50m..Each are moving with relativistic speeds 0.8c and 0.6c respectively in the same direction w.r.to a stationary Observer S.At t=0,the spaceship A is just behind the spaceship B..Find the time taken by the spaceship A???

[@sambichagwa1119]

8 years in and still helping us

[@MichelvanBiezen]

As long as students find the videos helpful and we are able, we'll keep making them.

[@iiscdiaries]

if the question well be the velocity of separation seen by an observer in the earth? then I guess we can't say something because we have to choose one spaceship as an event necessarily but when you are asking for say something about the observer on the earth we are asking to observe both spaceships means two events at the same time. this makes no sense? is that thinking right? please correct me

[@MichelvanBiezen]

The key to these types of problems is this: There are two observers. One observer (A) is on a stationary platform. A second observer is on a moving platform (B). And then there is an event which could be on the stationary platform, the moving platform, or move independently from either of the platforms.

[@iiscdiaries]

@@MichelvanBiezen OK thanks sir 😊👍

[@tryhardz-mc6089]

amazing explaintion

[@MichelvanBiezen]

It is an amazing universe out there.

[@shameershami7724]

Masha Allah amazing lecture sir

[@sutopamodak6247]

Hey if i send mail a Problem which i did in your way but having a negative sign problem Have you any problem??

[@MichelvanBiezen]

Make sure you make the proper distinction between observer A (in the stationary frame), observer B in the moving frame, and the experiment or item being observed. We have several examples of those and see if you can determine it by watching all the examples.

[@rajuroy5018]

Plz create more videos on solving problems on relativity sir!!

[@MichelvanBiezen]

There is a whole playlist. Did you see all of them?

[@danahjynetangayan2064]

5 hours before my exam. Thank you good sir

[@MichelvanBiezen]

Good luck in your exam.

[@xyliahvortex661]

I don’t get why we can’t take v’ to be -0.6 and plug it into the equation for v.

[@MichelvanBiezen]

When traveling at high velocities, our regular equations of kinematics break down. They are no longer applicable, since time progresses at different rates depending on the velocity an object is traveling at.

[@Therock151214]

Great Explanation absolutely amazing, thank you sir

[@MichelvanBiezen]

Glad you liked it.

[@gregorygoulenko3163]

van biezen you are awesome !

[@MichelvanBiezen]

Thanks!

[@manveitsok4990]

Can we solve that question in same way if relative speed ask.

[@MichelvanBiezen]

yes, but you will need to rearrange the equation to solve for the variable you want.

[@philipthomas982]

Why is it necessary to multiply both the numerator and denominator by - 1?

[@magembefrank2329]

Wow So Great

[@MichelvanBiezen]

Glad you liked it.

[@lilmiscuti3]

This is amazing!!!

[@HeatherSpoonheim]

Thanks so much. I plugged in -c for the velocity of the event, wanting to verify that light would appear to be going the speed of light even when you are going towards the beam of light at high velocity and guess what I got for a result? -c! Fun fun.

[@HeatherSpoonheim]

Heather Spoonheim And just checked with c for event velocity and got c - lovely, :D

[@MichelvanBiezen]

Heather Spoonheim That is great. That is how you get the appreciation of the concepts and the equations, by trying different things like that.

[@kacperkinastowski5583]

Nice bowtie, and very good lecture! You helped me a lot, thank you!

[@helmer5667]

That is insane

[@MichelvanBiezen]

Yes, I had trouble believing this was real when I was a student. 🙂

[@vijayavula4410]

What of they are moving perpendicularly?can you please comment?

[@MichelvanBiezen]

The effects of relativity act in the direction of travel or along the line of sight.

[@vijayavula4410]

@@MichelvanBiezen Thank u sir.i got it.

[@moedgerveni9075]

lifesaver!!

[@jasonking1284]

1.4c

[@vibhorbhatnagar7074]

sir please explain the numericals on relativistic addition where one particle moves in x axis another in y axis and v have to found the velocity of one relative to another

[@desertpointshacks6299]

What would be the combined velocity with which the two spaceships are approaching each other as viewed from Earth? Does that question make sense in relativistic physics? I.e in classical physics it would jst be the two object’s velocities added together (if they are travelling towards each other).

[@MichelvanBiezen]

Yes, it is very different at relativistic speeds.

[@desertpointshacks6299]

@@MichelvanBiezen Thank you very much :) You really are an excellent teacher!

[@iiscdiaries]

than what will be answer?

[@kristianbojinov6715]

@@iiscdiaries In classical physics with the 2 objects moving towards each other with speeds x and y, the combined relative speed would just be x+y

[@ironuranium3927]

if both u and v equal to the velocity of light (with respect to stationary reference frame) and they both travel at the positive x-direction then v(prime) equal to infinity but how this is possible nothing can move faster than light please say something about that?

[@masudrana5886]

❤️❤️❤️❤️❤️❤️❤️

[@ericsu4667]

The velocity transformation depends solely on the acceleration that seperates two inertial reference frames apart. Velocity transformation is independent of the speed of light. sites.google.com/view/physics-news/home/transformation

[@lawrence1318]

Using Relativistic equations doesn't prove Relativity is correct. All it does is beg the question. If the 2 ships are separated by a distance of 1.4 light seconds, they will collide with each other in 1 second. So they are necessarily traveling at 1.4c relative to each other. This is just a fact that is so obvious it beggars belief anyone could think otherwise. Similarly, if earth were to head toward a star, the star's light would reach the earth in less time than if the earth were to remain stationary. So the earth and the star's light are approaching each other at a speed greater than c and as evidence of this the light would be blue-shifted in the eyes of its observers on earth when it reaches the earth. Yes the light's speed is itself still c regardless of whether blue-shifted or not and this simply because its speed is equal to its wavelength multiplied by its frequency, but the *_extent_* to which it is blue-shifted in the eyes of those who see it will tell them how much faster than c its speed relative to them actually is. Relativity is ultimately based on Einstein's analysis of simultaneity .... on what is referred to as the Relativity of Simultaneity. At the following link I refute the Relativity of Simultaneity: vixra.org/pdf/2002.0103v1.pdf