Physics 62 Special Relativity (9 of 43) Relativistic Energy: A General Approach

[@valeriereid2337]

Thanks so much for making this channel very reliable.

[@MichelvanBiezen]

Glad you found this helpful

[@HvatrexHamam]

you are great teacher , now i understand the equations of relativity :) thanks a lot

[@Abhijitdas8710]

Loved the way of solving problems 😍

[@youssefrefaat1594]

Sir, please answers this question. the kinetic energy we just found is object's real KE or it is the relative one according to the observer? I guess it should be the relative one. Right? If we are doing an experiment and we would like to calculate the object's kinetic energy, which kinetic energy we will use? (suppose that the object is so fast) will we use the relative one ? which can be calculated by Einstein's equation? or we will use the classical method ( .5m v ^2)?

[@MichelvanBiezen]

The KE is typically measured by an observer at rest and thus the relativistic kinetic energy is what is measured, observed, and used in all interactions. We only use KE = (1/2) mv^2 for object moving at NON-relativistic speeds.

[@youssefrefaat1594]

But I still don't get it. Let me give you an example. Imagine a person moving with an extremely fast speed. We are trying to measure his kinetic energy. Thus, we will use Einstein's equations. Is the value we will get is the same as the value the person who is moving very fast will get? suppose that the person can measure his velocity. If this person collides elastically with another object and transfers all of his energy, the energy he will transfer ( measured by him) will be different than the value measured by us. I think the real (or I should say the "correct") value is the one the person measures not the one we (as observers) meaure

[@MichelvanBiezen]

When you are in a stationary lab and a particle moving very fast comes in and collides with a target and then you measure the energy transferred it will be equal to the relativistic KE of the first particle.

[@youssefrefaat1594]

now it is clear, thanks sir

[@TriviaQuizocity]

I would like to know how to aply this to the equasion for sintrifical force "(v^2)/r"?

[@MichelvanBiezen]

The equation for the centripetal force = m v^2 / R (centripetal acceleration = v^2 / R). It is the force required to keep a object moving in a circle with velocity v and radius R. The centrifugal force is the "apparent" force you appear to feel when you travel in a circle and the equation is the same. Using the "centrifugal" force often makes it easier to envision the situation and to solve the problems.

[@merveyurduseven6725]

both E_T=mc^2 and E_0=m_0c^2 are the equations at rest. but in the question, our mass is not at rest which we should use E^2=(pc)^2+(mc^2)^2. why did we use rest energy sir? @michelvanbiezen

[@zakirhussain-js9ku]

In E=MC2, mass & energy are equivalent & mutually convertible as C is constant. By same logic in non-relativistic & relativistic kinetic energy mass is constant, therefore Energy & motion must be equivalent & mutually convertible. Equvalence of energy to both mass and motion implies that mass & motion are also equivalent and mutually convertible. In annihilation of stationary electron and positron mass is converted to motion of photons. In pair production motion of photon is converted back to mass. In nuclear reactions mass is converted to motion of neutrons and gamma ray photons release during reaction. Energy, Mass and motion are equivalent & mutually convertible.

[@MichelvanBiezen]

Thank you for the input. 🙂

[@zakirhussain-js9ku]

Since the gravitational field of moving mass does not increase, it is said that relativistic mass is not real. This suggest that real & usable kinetic energy of a moving mass is only 1/2mv² where m is rest or real mass. Relativistic mass kinetic energy ( 1.392- 1.125)10¹⁶J may be dissipating into Space as unrecoverable energy.

[@MichelvanBiezen]

"it is said" who is saying that?

[@pramilarathore324]

Amazing explanation ☺️

[@MichelvanBiezen]

Thank you. Glad you found our videos! 🙂

[@blackbell1478]

Please help me answer this question Given that the mass of electron =9×10-³¹kg and that the velocity of light is =3×108 m5¹ Calculate the following (A) the kinetic energy of a gas molecule of mass 5×10²⅝kg moving the velocity of 400ms¹ (B) the kinetic energy of an electron moving with an electron 2×10 ms¹ (C) the energy of a quantum of green light of wave length 55onm (D) the energy of a quantum of ultraviolet radiation by wave length 100nm (2) a quantum of energy 4.5 election volt is incident on a calcum surface what is the energy of the emitted election (3) what is the velocity of an electron of energy 2.5 ev

[@ajob_kaktarua]

You are really an awsome teacher Sir! For the last 2 years whenever I coudn't understand any topic, you are the last resource when everyone else failed :) Sir , according to 6:40, if an object moves with more than 0.1c velocity, E(k)=1/2mv^2 wont work. But in the case of x-ray tube electron, in all the examples though v>0.1c, still "eV=1/2mv^2" is used in my college book. Is it wrong? Also, if I want to use relatvity here, how can I use that when only information given in the qs is the volt difference in between x-ray tube (velocity of electron is asked to be found). Thanks :)

[@MichelvanBiezen]

Thank you and here is the answer to your question. At 0.1c, the relativity effects are still very small and can be ignored (the error would be 0.5%) At 0.5c the error would be 13.5%. So in most cases the velocities are well below 0.5c and the effect can be ignored. (off course if you want to be accurate, it is advised to use these effects in the calculations)

[@simranagrawal136]

Does every rest mass(for eg,humans) have rest mass energy? And if yes how are we able to contain that much energy inside us?

[@MichelvanBiezen]

It is indeed one of the great mysteries of the universe.

[@sparkstarter]

I notice that the difference is only about 20% could one make a table of correction factors to be used with the traditional KE = 1/2 m v^2 ? The benefit of this would be that if somebody wanted to play with the concepts of kinetic energy at near light speed, they wouldn't need to remember the relativistic derivation, they could just apply a correction factor to the old KE equation, or would they be missing something?

[@marksteven2716]

I don't think a table is necessary since the ''correction factor'' is already gamma itself. Given that it is pretty simple to apply each time, a programmer could easily build a model using this fact.

[@redwolf6334]

very good video thanks

[@MichelvanBiezen]

Glad you liked it!

[@sanjidislam3825]

Einstein's theory of relativity stipulate that the work required to assemble a charge is stored as energy in the mass and is equal to mc2, where m is the mass and c=3*108 (m/s) is the velocity of light. Assuming the electron to be a perfect sphere, find its radius from its charge and mass (9.1*10-31 kg). Pls solve it

[@MichelvanBiezen]

Not sure what you mean by: "find its radius". Can you explain what you mean by "it"? Are you asking to find the radius of the electron, given the charge and mass of the electron?

[@satyamkumar567]

Sir please make complete video on general theory of relativity

[@MichelvanBiezen]

We have some videos on general relativity topics in our quantum mechanics playlists, but you suggestion is a good one, we should put a playlist together on that.

[@Joselo09111]

Thank you!

[@MichelvanBiezen]

You're welcome!

[@oussamabaadj9355]

Hi sir, i have to study these subjects : atomic physics, particle physics, quantum mechanics and relativity. could you please classify them for me, because i do not know from where i begin.

[@MichelvanBiezen]

The listings are listed out in this video: How to Find the Playlists on Physics (3 of 3) E&M, Optics, Modern Physics

[@oussamabaadj9355]

Thank you sir

[@rdrahul9444]

Why 1/2 (rMo)v^2 doesn't give the right answer?

[@MichelvanBiezen]

Start with the definition of KE at the relativistic level.

[@amirnawaz6618]

New way to solve problem great I will copy you after today sir?? If u allowed me..

[@MichelvanBiezen]

No problem. Glad you found our videos. 🙂

[@jasimmathsandphysics]

Why is the total energy equal to mc^2? Why not mv^2/2?

[@MichelvanBiezen]

(1/2) mv^2 represent kinetic enrgy. E = mc^2 is the energy released when mass is converted to energy. They are very different concepts.

[@jasimmathsandphysics]

@@MichelvanBiezen so when an objects travels at relativistic speeds, its mass is converted into energy?

[@MichelvanBiezen]

When an object travels at relativistic speed, it appears to gain mass. The energy must come from an outside source, not from the object itself.

[@sakshamgulati1578]

awesome video!!!

[@honeyx7879]

thank you so much!

[@diyrobotronics]

again thank u sir

[@MichelvanBiezen]

You are welcome. 🙂

[@ericsu4667]

E is not mc2. Two Eurpean synchrotrons use experimental data to prove that energy is not mc2. Detail in ™83 Representation of Energy in Synchrotron™ on sites.google.com/view/physics-news/home/updates

[@olasek7972]

What a bunch of sorry idiots, and those idiots actually think someone cares about their version of “science”.

[@User-ei2kw]

sir relativistic mass does not exist, mass is constant

[@MichelvanBiezen]

Essentially you are probably correct. Even though all the equations work out when we consider that there is relativistic mass and all the particles act as if they have relativistic mass, the actual mechanism is probably related to momentum and the motion through space

[@masudrana5886]

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